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P6696 [BalticOI 2020] 图 (Day2)
题目描述
你有一个无向图,每条边都有一种颜色:红或者黑。
你要做的就是为每个节点配一个实数点权,使得:
- 对于每条黑色边,两个端点的点权之和为 1 1 1
- 对于每条红色边,两个端点的点权之和为 2 2 2
- 所有点权的绝对值之和是最小的
求一种点权的分配方案。
输入格式
第一行两个整数 N , M N,M N,M 代表点数和边数。
所有点的编号为 1 1 1 到 N N N。
接下来 M M M 行每行三个整数 a , b , c a,b,c a,b,c 描述一条边端点为 a a a 和 b b b,如果 c c c 是 1 1 1 那么这条边是黑色边,如果 c c c 是 2 2 2 那么这条边是红色边。
输出格式
如果有解,首先第一行输出 YES
,然后第二行 N N N 个整数代表可能的一组点权。
多组解输出任意一组即可。
如果无解,一行一个字符串 NO
。
输入输出样例 #1
输入 #1
4 4
1 2 1
2 3 2
1 3 2
3 4 1
输出 #1
YES
0.5 0.5 1.5 -0.5
输入输出样例 #2
输入 #2
2 1
1 2 1
输出 #2
YES
0.3 0.7
输入输出样例 #3
输入 #3
3 2
1 2 2
2 3 2
输出 #3
YES
0 2 0
输入输出样例 #4
输入 #4
3 4
1 2 2
2 2 1
2 1 1
1 2 2
输出 #4
NO
说明/提示
评测方式
您的输出被评判为正确,当且仅当:
- 每条边所连两点的点权和与该边要求的点权间的误差不超过 1 0 − 6 10^{-6} 10−6。
- 所有点权的绝对值之和与标准答案误差不超过 1 0 − 6 10^{-6} 10−6。
数据规模与约定
本题采用捆绑测试。
- Subtask 1(5 pts): N ≤ 5 N \le 5 N≤5, M ≤ 14 M \le 14 M≤14。
- Subtask 2(12 pts): N ≤ 100 N \le 100 N≤100。
- Subtask 3(17 pts): N ≤ 1000 N \le 1000 N≤1000。
- Subtask 4(24 pts): N ≤ 1 0 4 N \le 10^4 N≤104。
- Subtask 5(42 pts):无特殊限制。
对于 100 % 100\% 100% 的数据, 1 ≤ N ≤ 1 0 5 1 \le N \le 10^5 1≤N≤105, 0 ≤ M ≤ 2 × 1 0 5 0 \le M \le 2 \times 10^5 0≤M≤2×105。
本题使用 Special Judge。
[BalticOI 2020] 图 (Day2) DFS|普及+
注意 :本题有自环和重边,如样例4。自环可以按奇数环处理。重边的权重必须同。可排序后比较。
性质一 :不同的连通区域互不影响。各连通区域要分别处理。
本题等效于,黑色边之和2,红边之和4,最终结果除以2。
如果没有环,任意节点为x,然后计算其它节点的点权。下面讨论环:
长度为3的奇数环,A的点权是x1,AB、BC、CA的边权分别为ew1,ew2,ew2。x2 = ew1-x1。x3=ew2-(ew1-x1)=x1+ew2-ew1。x3+x1=ew3。即:
2x1= ew3-ew2+ew1。即: x = (ew3+ew1-ew2)/2。奇数环类似。
长度为4的偶数环:x4=ew3-x3 = ew3-ew2+ew1-x1。x4+x1 =ew4,即ew4-ew3+ew2-ew1=0。
结论一:偶数环,边权之和必须为0。奇数环可以确定。
DFS
DFS时,任意节点只在第一次DFS时访问临接点,即每条边只访问一次。故时间复杂度:O(n+m)。
DFS(cur, vGP,is) cur是当前节点,vGP是cur的祖先,vGP.back()是cur的父节点。vw[i]记录gp[i]到父节点的权, is[i] 记录i是否是cur的祖先。如果is[cur]成立,说明遇到环(不是第一次访问), 不访问临接点。
如果是偶数环,判断边权是否是0,如果不是0,直接结束程序,无解。如果是奇数环,求出cur的点权。每个环只会访问两次,cur是第一个DFS到的点,访问两次是无向边。如果有奇数环,相互矛盾则某个节点两次计算的值不相同。
BFS
如果某个连通区域有奇数环,则直接计算各点权。如果有连通区域没有奇数环,则任意一点权为x,其它点可以kx+b, k是1或-1。|x+b|等效于|x-(-b)|,|-x+b|等效于|x-b| 。我们要想各点权绝对值之和最小,就是x到bs各点距离和最小。 x取bs的中位数。如果b有偶数中则中间任意一个。k为1,bs增加-b;k为-1,bs增加b。求出x后,计算本连通区域各点的值。
代码:每个测试点用少量样例过不了,过于复杂很难排查
cpp
#include <iostream>
#include <sstream>
#include <vector>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<string>
#include<algorithm>
#include<functional>
#include<queue>
#include <stack>
#include<iomanip>
#include<numeric>
#include <math.h>
#include <climits>
#include<assert.h>
#include<cstring>
#include<list>
#include <bitset>
using namespace std;
template<class T1, class T2>
std::istream& operator >> (std::istream& in, pair<T1, T2>& pr) {
in >> pr.first >> pr.second;
return in;
}
template<class T1, class T2, class T3 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3>& t) {
in >> get<0>(t) >> get<1>(t) >> get<2>(t);
return in;
}
template<class T1, class T2, class T3, class T4 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3, T4>& t) {
in >> get<0>(t) >> get<1>(t) >> get<2>(t) >> get<3>(t);
return in;
}
template<class T = int>
vector<T> Read() {
int n;
cin >> n;
vector<T> ret(n);
for (int i = 0; i < n; i++) {
cin >> ret[i];
}
return ret;
}
template<class T = int>
vector<T> ReadNotNum() {
vector<T> ret;
T tmp;
while (cin >> tmp) {
ret.emplace_back(tmp);
if ('\n' == cin.get()) { break; }
}
return ret;
}
template<class T = int>
vector<T> Read(int n) {
vector<T> ret(n);
for (int i = 0; i < n; i++) {
cin >> ret[i];
}
return ret;
}
template<int N = 1'000'000>
class COutBuff
{
public:
COutBuff() {
m_p = puffer;
}
template<class T>
void write(T x) {
int num[28], sp = 0;
if (x < 0)
*m_p++ = '-', x = -x;
if (!x)
*m_p++ = 48;
while (x)
num[++sp] = x % 10, x /= 10;
while (sp)
*m_p++ = num[sp--] + 48;
AuotToFile();
}
void writestr(const char* sz) {
strcpy(m_p, sz);
m_p += strlen(sz);
AuotToFile();
}
inline void write(char ch)
{
*m_p++ = ch;
AuotToFile();
}
inline void ToFile() {
fwrite(puffer, 1, m_p - puffer, stdout);
m_p = puffer;
}
~COutBuff() {
ToFile();
}
private:
inline void AuotToFile() {
if (m_p - puffer > N - 100) {
ToFile();
}
}
char puffer[N], * m_p;
};
template<int N = 1'000'000>
class CInBuff
{
public:
inline CInBuff() {}
inline CInBuff<N>& operator>>(char& ch) {
FileToBuf();
ch = *S++;
return *this;
}
inline CInBuff<N>& operator>>(int& val) {
FileToBuf();
int x(0), f(0);
while (!isdigit(*S))
f |= (*S++ == '-');
while (isdigit(*S))
x = (x << 1) + (x << 3) + (*S++ ^ 48);
val = f ? -x : x; S++;//忽略空格换行
return *this;
}
inline CInBuff& operator>>(long long& val) {
FileToBuf();
long long x(0); int f(0);
while (!isdigit(*S))
f |= (*S++ == '-');
while (isdigit(*S))
x = (x << 1) + (x << 3) + (*S++ ^ 48);
val = f ? -x : x; S++;//忽略空格换行
return *this;
}
template<class T1, class T2>
inline CInBuff& operator>>(pair<T1, T2>& val) {
*this >> val.first >> val.second;
return *this;
}
template<class T1, class T2, class T3>
inline CInBuff& operator>>(tuple<T1, T2, T3>& val) {
*this >> get<0>(val) >> get<1>(val) >> get<2>(val);
return *this;
}
template<class T1, class T2, class T3, class T4>
inline CInBuff& operator>>(tuple<T1, T2, T3, T4>& val) {
*this >> get<0>(val) >> get<1>(val) >> get<2>(val) >> get<3>(val);
return *this;
}
template<class T = int>
inline CInBuff& operator>>(vector<T>& val) {
int n;
*this >> n;
val.resize(n);
for (int i = 0; i < n; i++) {
*this >> val[i];
}
return *this;
}
template<class T = int>
vector<T> Read(int n) {
vector<T> ret(n);
for (int i = 0; i < n; i++) {
*this >> ret[i];
}
return ret;
}
template<class T = int>
vector<T> Read() {
vector<T> ret;
*this >> ret;
return ret;
}
private:
inline void FileToBuf() {
const int canRead = m_iWritePos - (S - buffer);
if (canRead >= 100) { return; }
if (m_bFinish) { return; }
for (int i = 0; i < canRead; i++)
{
buffer[i] = S[i];//memcpy出错
}
m_iWritePos = canRead;
buffer[m_iWritePos] = 0;
S = buffer;
int readCnt = fread(buffer + m_iWritePos, 1, N - m_iWritePos, stdin);
if (readCnt <= 0) { m_bFinish = true; return; }
m_iWritePos += readCnt;
buffer[m_iWritePos] = 0;
S = buffer;
}
int m_iWritePos = 0; bool m_bFinish = false;
char buffer[N + 10], * S = buffer;
};
class KMP
{
public:
virtual int Find(const string& s, const string& t)
{
CalLen(t);
for (int i1 = 0, j = 0; i1 < s.length(); )
{
for (; (j < t.length()) && (i1 + j < s.length()) && (s[i1 + j] == t[j]); j++);
//i2 = i1 + j 此时s[i1,i2)和t[0,j)相等 s[i2]和t[j]不存在或相等
//t[0,j)的结尾索引是j-1,所以最长公共前缀为m_vLen[j-1],简写为y 则t[0,y)等于t[j-y,j)等于s[i2-y,i2)
if (0 == j)
{
i1++;
continue;
}
const int i2 = i1 + j;
j = m_vLen[j - 1];
i1 = i2 - j;//i2不变
}
return -1;
}
//vector<int> m_vSameLen;//m_vSame[i]记录 s[i...]和t[0...]最长公共前缀,增加可调试性 部分m_vSameLen[i]会缺失
//static vector<int> Next(const string& s)
//{// j = vNext[i] 表示s[0,i]的最大公共前后缀是s[0,j]
// const int len = s.length();
// vector<int> vNext(len, -1);
// for (int i = 1; i < len; i++)
// {
// int next = vNext[i - 1];
// while ((-1 != next) && (s[next + 1] != s[i]))
// {
// next = vNext[next];
// }
// vNext[i] = next + (s[next + 1] == s[i]);
// }
// return vNext;
//}
const vector<int> CalLen(const string& str)
{
m_vLen.resize(str.length());
for (int i = 1; i < str.length(); i++)
{
int next = m_vLen[i - 1];
while (str[next] != str[i])
{
if (0 == next)
{
break;
}
next = m_vLen[next - 1];
}
m_vLen[i] = next + (str[next] == str[i]);
}
return m_vLen;
}
protected:
int m_c;
vector<int> m_vLen;//m_vLen[i] 表示str[0,i]的最长公共前后缀的长度
};
template<long long MOD = 1000000007, class T1 = int, class T2 = long long>
class C1097Int
{
public:
C1097Int(T1 iData = 0) :m_iData(iData% MOD)
{
}
C1097Int(T2 llData) :m_iData(llData% MOD) {
}
C1097Int operator+(const C1097Int& o)const
{
return C1097Int(((T2)m_iData + o.m_iData) % MOD);
}
C1097Int& operator+=(const C1097Int& o)
{
m_iData = ((T2)m_iData + o.m_iData) % MOD;
return *this;
}
C1097Int& operator-=(const C1097Int& o)
{
m_iData = ((T2)MOD + m_iData - o.m_iData) % MOD;
return *this;
}
C1097Int operator-(const C1097Int& o)
{
return C1097Int(((T2)MOD + m_iData - o.m_iData) % MOD);
}
C1097Int operator*(const C1097Int& o)const
{
return((T2)m_iData * o.m_iData) % MOD;
}
C1097Int& operator*=(const C1097Int& o)
{
m_iData = ((T2)m_iData * o.m_iData) % MOD;
return *this;
}
C1097Int operator/(const C1097Int& o)const
{
return *this * o.PowNegative1();
}
C1097Int& operator/=(const C1097Int& o)
{
*this /= o.PowNegative1();
return *this;
}
bool operator==(const C1097Int& o)const
{
return m_iData == o.m_iData;
}
bool operator<(const C1097Int& o)const
{
return m_iData < o.m_iData;
}
C1097Int pow(T2 n)const
{
C1097Int iRet = (T1)1, iCur = *this;
while (n)
{
if (n & 1)
{
iRet *= iCur;
}
iCur *= iCur;
n >>= 1;
}
return iRet;
}
C1097Int PowNegative1()const
{
return pow(MOD - 2);
}
T1 ToInt()const
{
return ((T2)m_iData + MOD) % MOD;
}
private:
T1 m_iData = 0;;
};
class CParentToNeiBo
{
public:
CParentToNeiBo(const vector<int>& parents)
{
m_vNeiBo.resize(parents.size());
for (int i = 0; i < parents.size(); i++)
{
if (-1 == parents[i])
{
m_root = i;
}
else
{
m_vNeiBo[parents[i]].emplace_back(i);
}
}
}
vector<vector<int>> m_vNeiBo;
int m_root = -1;
};
class CBFSLeve {
public:
static vector<int> Leve(const vector<vector<int>>& neiBo, vector<int> start) {
vector<int> leves(neiBo.size(), -1);
for (const auto& s : start) {
leves[s] = 0;
}
for (int i = 0; i < start.size(); i++) {
for (const auto& next : neiBo[start[i]]) {
if (-1 != leves[next]) { continue; }
leves[next] = leves[start[i]] + 1;
start.emplace_back(next);
}
}
return leves;
}
template<class NextFun>
static vector<int> Leve(int N, NextFun nextFun, vector<int> start) {
vector<int> leves(N, -1);
for (const auto& s : start) {
leves[s] = 0;
}
for (int i = 0; i < start.size(); i++) {
auto nexts = nextFun(start[i]);
for (const auto& next : nexts) {
if (-1 != leves[next]) { continue; }
leves[next] = leves[start[i]] + 1;
start.emplace_back(next);
}
}
return leves;
}
static vector<vector<int>> LeveNodes(const vector<int>& leves) {
const int iMaxLeve = *max_element(leves.begin(), leves.end());
vector<vector<int>> ret(iMaxLeve + 1);
for (int i = 0; i < leves.size(); i++) {
ret[leves[i]].emplace_back(i);
}
return ret;
};
static vector<int> LeveSort(const vector<int>& leves) {
const int iMaxLeve = *max_element(leves.begin(), leves.end());
vector<vector<int>> leveNodes(iMaxLeve + 1);
for (int i = 0; i < leves.size(); i++) {
leveNodes[leves[i]].emplace_back(i);
}
vector<int> ret;
for (const auto& v : leveNodes) {
ret.insert(ret.end(), v.begin(), v.end());
}
return ret;
};
};
class CParents
{
public:
CParents(vector<int>& vParent, long long iMaxDepth)
{
int iBitNum = 0;
for (; iMaxDepth; iBitNum++) {
const auto mask = 1LL << iBitNum;
if (mask & iMaxDepth) { iMaxDepth = iMaxDepth ^ mask; }
}
const int n = vParent.size();
m_vParents.assign(iBitNum + 1, vector<int>(n, -1));
m_vParents[0] = vParent;
//树上倍增
for (int i = 1; i < m_vParents.size(); i++)
{
for (int j = 0; j < n; j++)
{
const int iPre = m_vParents[i - 1][j];
if (-1 != iPre)
{
m_vParents[i][j] = m_vParents[i - 1][iPre];
}
}
}
}
int GetParent(int iNode, int iDepth)const
{
int iParent = iNode;
for (int iBit = 0; iBit < m_vParents.size(); iBit++)
{
if (-1 == iParent)
{
return iParent;
}
if (iDepth & (1 << iBit))
{
iParent = m_vParents[iBit][iParent];
}
}
return iParent;
}
inline int GetBitCnt()const { return m_vParents.size(); };
inline const int& GetPow2Parent(int iNode, int n)const {
return m_vParents[n][iNode];
}
//在leftNodeExclude的1到rightLeve级祖先中查找符合pr的最近祖先
template<class _Pr>
int FindFirst(int leftNodeExclude, int rightLeve, _Pr pr) {
for (int iBit = GetBitCnt() - 1; iBit >= 0; iBit--) {
const int iMask = 1 << iBit;
if (!(iMask & rightLeve)) { continue; }
if (pr(m_vParents[iBit][leftNodeExclude])) {
return BFindFirst(leftNodeExclude, iBit, pr);
}
leftNodeExclude = m_vParents[iBit][leftNodeExclude];
}
return -1;
}
protected:
//在leftNodeExclude的1到2^pow^级祖先中查找符合pr的最近祖先
template<class _Pr>
int BFindFirst(int leftNodeExclude, int pow, _Pr pr) {
while (pow > 0) {
const int& mid = m_vParents[pow - 1][leftNodeExclude];
if (pr(mid)) {
}
else {
leftNodeExclude = mid;
}
pow--;
}
return m_vParents[0][leftNodeExclude];
}
vector<vector<int>> m_vParents;
};
class CNeiBo
{
public:
static vector<vector<int>> Two(int n, const vector<pair<int, int>>& edges, bool bDirect, int iBase = 0)
{
vector<vector<int>> vNeiBo(n);
for (const auto& [i1, i2] : edges)
{
vNeiBo[i1 - iBase].emplace_back(i2 - iBase);
if (!bDirect)
{
vNeiBo[i2 - iBase].emplace_back(i1 - iBase);
}
}
return vNeiBo;
}
static vector<vector<int>> Two(int n, const vector<vector<int>>& edges, bool bDirect, int iBase = 0)
{
vector<vector<int>> vNeiBo(n);
for (const auto& v : edges)
{
vNeiBo[v[0] - iBase].emplace_back(v[1] - iBase);
if (!bDirect)
{
vNeiBo[v[1] - iBase].emplace_back(v[0] - iBase);
}
}
return vNeiBo;
}
static vector<vector<std::pair<int, int>>> Three(int n, vector<vector<int>>& edges, bool bDirect, int iBase = 0)
{
vector<vector<std::pair<int, int>>> vNeiBo(n);
for (const auto& v : edges)
{
vNeiBo[v[0] - iBase].emplace_back(v[1] - iBase, v[2]);
if (!bDirect)
{
vNeiBo[v[1] - iBase].emplace_back(v[0] - iBase, v[2]);
}
}
return vNeiBo;
}
static vector<vector<std::pair<int, int>>> Three(int n, const vector<tuple<int, int, int>>& edges, bool bDirect, int iBase = 0)
{
vector<vector<std::pair<int, int>>> vNeiBo(n);
for (const auto& [u, v, w] : edges)
{
vNeiBo[u - iBase].emplace_back(v - iBase, w);
if (!bDirect)
{
vNeiBo[v - iBase].emplace_back(u - iBase, w);
}
}
return vNeiBo;
}
static vector<vector<int>> Mat(vector<vector<int>>& neiBoMat)
{
vector<vector<int>> neiBo(neiBoMat.size());
for (int i = 0; i < neiBoMat.size(); i++)
{
for (int j = i + 1; j < neiBoMat.size(); j++)
{
if (neiBoMat[i][j])
{
neiBo[i].emplace_back(j);
neiBo[j].emplace_back(i);
}
}
}
return neiBo;
}
};
class Solution {
public:
vector<float> Ans(int N, const vector<tuple<int, int, int>>& edge) {
m_iN = N;
m_neiBo = CNeiBo::Three(N, edge, false, 1);
for (auto& v : m_neiBo) {
for (auto& [next, w] : v) {
w *= 2;
}
sort(v.begin(), v.end());
for (int i = 1; i < v.size(); i++) {
if ((v[i].first == v[i - 1].first) && (v[i].second != v[i - 1].second)) { return {}; }
}
}
m_vk.assign(N, -100); m_vb.assign(N, 0);
for (int i = 0; i < N; i++)
{
if (-100 != m_vk[i]) { continue; }
int ring = -1;
if (!DFS(ring, i)) {
return {};
}
if (-1 == ring) {
ring = i; m_vk[ring] = 1;
}
BFS(ring);
}
vector<float> ans;
for (int i = 0; i < N; i++) {
ans.emplace_back(m_vb[i] / 2.0);
}
return ans;
}
bool DFS(int& ring, int cur) {
vector<int> gp, vw; vector<bool> is(m_iN);
function<bool(int, int)> DFS = [&](int cur, int w) {
if (is[cur]) {//遇到环
ring = cur;
int sign = -1, sum = w;
for (int i = gp.size() - 1; cur != gp[i]; i--) {
sum += sign * vw[i];
sign = (1 == sign) ? -1 : 1;
}
if ((1 == sign) && (sum != 0)) {
return false;
}
if (-1 == sign) {
if ((-100 != m_vk[cur]) && (2 * m_vb[cur] != sum)) {
return false;
}
m_vk[cur] = 0; m_vb[cur] = sum / 2;
}
return true;
}
gp.emplace_back(cur); vw.emplace_back(w); is[cur] = true;
for (const auto& [next, w] : m_neiBo[cur]) {
if ((gp.size() >= 2) && (gp[gp.size() - 2] == next)) { continue; }
if (!DFS(next, w)) { return false; }
}
gp.pop_back(); vw.pop_back(); is[cur] = false;
return true;
};
return DFS(cur, 0);
}
void BFS(int root) {
queue<int> que; vector<bool> vis(m_iN);
vector<pair<int, int>> bsNode;
que.emplace(root); vis[root] = true;
while (que.size()) {
auto cur = que.front(); que.pop();
if (1 == m_vk[cur]) {
bsNode.emplace_back(-m_vb[cur], cur);
}
else if (-1 == m_vk[cur]) {
bsNode.emplace_back(m_vb[cur], cur);
}
for (const auto& [next, w] : m_neiBo[cur]) {
m_vk[next] = -m_vk[cur];
m_vb[next] = w - m_vb[cur];
if (!vis[next]) {
que.emplace(next); vis[next] = true;
}
}
}
if (bsNode.empty()) { return; }
nth_element(bsNode.begin(), bsNode.begin() + bsNode.size() / 2, bsNode.end());
const auto x = bsNode[bsNode.size() / 2].first;
for (const auto& [tmp, cur] : bsNode) {
m_vb[cur] += m_vk[cur] * x;
}
}
int m_iN;
vector<int> m_vk, m_vb;
vector<vector<std::pair<int, int>>> m_neiBo;
};
int main() {
#ifdef _DEBUG
freopen("a.in", "r", stdin);
#endif // DEBUG
ios::sync_with_stdio(0); cin.tie(nullptr);
int n;
cin >> n;
auto edge = Read<tuple<int, int, int>>();
#ifdef _DEBUG
printf("N=%d",n);
//cout << ",s=" << s;
Out(edge, ",edge=");
//Out(ws, ",hs=");
//Out(que, ",que=");
/*Out(que, "que=");*/
#endif // DEBUG
auto res = Solution().Ans(n,edge);
cout << (res.size()?"YES":"NO" )<< '\n';
for (const auto& f : res) {
cout << f << " ";
}
return 0;
}
简化
每个连通块,直接令第一个是x,计算其它点。
如果每个点第一次时k1x+b1,第二次是k2x+b2。
如果: k 1 = = k 2 , b 1 = = b 2 k1 == k2,b1 == b2 k1==k2,b1==b2, 忽略。
如果: k 1 = = k 2 , b 1 ≠ b 2 k1 == k2, b1 \neq b2 k1==k2,b1=b2 ,出错。
如果: k 1 ≠ k 2 k1 \neq k2 k1=k2 ,则 x = b 2 − b 1 k 1 − k 2 x= \frac{b2-b1}{k1-k2} x=k1−k2b2−b1
如果没有计算出x,则通同过中位数贪心计算x。
最后将计算的结果演算判断是否无解。中间可以不判断非法,最后统一判断。
核心代码
cpp
#include <iostream>
#include <sstream>
#include <vector>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<string>
#include<algorithm>
#include<functional>
#include<queue>
#include <stack>
#include<iomanip>
#include<numeric>
#include <math.h>
#include <climits>
#include<assert.h>
#include<cstring>
#include<list>
#include <bitset>
using namespace std;
template<class T1, class T2>
std::istream& operator >> (std::istream& in, pair<T1, T2>& pr) {
in >> pr.first >> pr.second;
return in;
}
template<class T1, class T2, class T3 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3>& t) {
in >> get<0>(t) >> get<1>(t) >> get<2>(t);
return in;
}
template<class T1, class T2, class T3, class T4 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3, T4>& t) {
in >> get<0>(t) >> get<1>(t) >> get<2>(t) >> get<3>(t);
return in;
}
template<class T = int>
vector<T> Read() {
int n;
cin >> n;
vector<T> ret(n);
for (int i = 0; i < n; i++) {
cin >> ret[i];
}
return ret;
}
template<class T = int>
vector<T> ReadNotNum() {
vector<T> ret;
T tmp;
while (cin >> tmp) {
ret.emplace_back(tmp);
if ('\n' == cin.get()) { break; }
}
return ret;
}
template<class T = int>
vector<T> Read(int n) {
vector<T> ret(n);
for (int i = 0; i < n; i++) {
cin >> ret[i];
}
return ret;
}
template<int N = 1'000'000>
class COutBuff
{
public:
COutBuff() {
m_p = puffer;
}
template<class T>
void write(T x) {
int num[28], sp = 0;
if (x < 0)
*m_p++ = '-', x = -x;
if (!x)
*m_p++ = 48;
while (x)
num[++sp] = x % 10, x /= 10;
while (sp)
*m_p++ = num[sp--] + 48;
AuotToFile();
}
void writestr(const char* sz) {
strcpy(m_p, sz);
m_p += strlen(sz);
AuotToFile();
}
inline void write(char ch)
{
*m_p++ = ch;
AuotToFile();
}
inline void ToFile() {
fwrite(puffer, 1, m_p - puffer, stdout);
m_p = puffer;
}
~COutBuff() {
ToFile();
}
private:
inline void AuotToFile() {
if (m_p - puffer > N - 100) {
ToFile();
}
}
char puffer[N], * m_p;
};
template<int N = 1'000'000>
class CInBuff
{
public:
inline CInBuff() {}
inline CInBuff<N>& operator>>(char& ch) {
FileToBuf();
ch = *S++;
return *this;
}
inline CInBuff<N>& operator>>(int& val) {
FileToBuf();
int x(0), f(0);
while (!isdigit(*S))
f |= (*S++ == '-');
while (isdigit(*S))
x = (x << 1) + (x << 3) + (*S++ ^ 48);
val = f ? -x : x; S++;//忽略空格换行
return *this;
}
inline CInBuff& operator>>(long long& val) {
FileToBuf();
long long x(0); int f(0);
while (!isdigit(*S))
f |= (*S++ == '-');
while (isdigit(*S))
x = (x << 1) + (x << 3) + (*S++ ^ 48);
val = f ? -x : x; S++;//忽略空格换行
return *this;
}
template<class T1, class T2>
inline CInBuff& operator>>(pair<T1, T2>& val) {
*this >> val.first >> val.second;
return *this;
}
template<class T1, class T2, class T3>
inline CInBuff& operator>>(tuple<T1, T2, T3>& val) {
*this >> get<0>(val) >> get<1>(val) >> get<2>(val);
return *this;
}
template<class T1, class T2, class T3, class T4>
inline CInBuff& operator>>(tuple<T1, T2, T3, T4>& val) {
*this >> get<0>(val) >> get<1>(val) >> get<2>(val) >> get<3>(val);
return *this;
}
template<class T = int>
inline CInBuff& operator>>(vector<T>& val) {
int n;
*this >> n;
val.resize(n);
for (int i = 0; i < n; i++) {
*this >> val[i];
}
return *this;
}
template<class T = int>
vector<T> Read(int n) {
vector<T> ret(n);
for (int i = 0; i < n; i++) {
*this >> ret[i];
}
return ret;
}
template<class T = int>
vector<T> Read() {
vector<T> ret;
*this >> ret;
return ret;
}
private:
inline void FileToBuf() {
const int canRead = m_iWritePos - (S - buffer);
if (canRead >= 100) { return; }
if (m_bFinish) { return; }
for (int i = 0; i < canRead; i++)
{
buffer[i] = S[i];//memcpy出错
}
m_iWritePos = canRead;
buffer[m_iWritePos] = 0;
S = buffer;
int readCnt = fread(buffer + m_iWritePos, 1, N - m_iWritePos, stdin);
if (readCnt <= 0) { m_bFinish = true; return; }
m_iWritePos += readCnt;
buffer[m_iWritePos] = 0;
S = buffer;
}
int m_iWritePos = 0; bool m_bFinish = false;
char buffer[N + 10], * S = buffer;
};
class KMP
{
public:
virtual int Find(const string& s, const string& t)
{
CalLen(t);
for (int i1 = 0, j = 0; i1 < s.length(); )
{
for (; (j < t.length()) && (i1 + j < s.length()) && (s[i1 + j] == t[j]); j++);
//i2 = i1 + j 此时s[i1,i2)和t[0,j)相等 s[i2]和t[j]不存在或相等
//t[0,j)的结尾索引是j-1,所以最长公共前缀为m_vLen[j-1],简写为y 则t[0,y)等于t[j-y,j)等于s[i2-y,i2)
if (0 == j)
{
i1++;
continue;
}
const int i2 = i1 + j;
j = m_vLen[j - 1];
i1 = i2 - j;//i2不变
}
return -1;
}
//vector<int> m_vSameLen;//m_vSame[i]记录 s[i...]和t[0...]最长公共前缀,增加可调试性 部分m_vSameLen[i]会缺失
//static vector<int> Next(const string& s)
//{// j = vNext[i] 表示s[0,i]的最大公共前后缀是s[0,j]
// const int len = s.length();
// vector<int> vNext(len, -1);
// for (int i = 1; i < len; i++)
// {
// int next = vNext[i - 1];
// while ((-1 != next) && (s[next + 1] != s[i]))
// {
// next = vNext[next];
// }
// vNext[i] = next + (s[next + 1] == s[i]);
// }
// return vNext;
//}
const vector<int> CalLen(const string& str)
{
m_vLen.resize(str.length());
for (int i = 1; i < str.length(); i++)
{
int next = m_vLen[i - 1];
while (str[next] != str[i])
{
if (0 == next)
{
break;
}
next = m_vLen[next - 1];
}
m_vLen[i] = next + (str[next] == str[i]);
}
return m_vLen;
}
protected:
int m_c;
vector<int> m_vLen;//m_vLen[i] 表示str[0,i]的最长公共前后缀的长度
};
template<long long MOD = 1000000007, class T1 = int, class T2 = long long>
class C1097Int
{
public:
C1097Int(T1 iData = 0) :m_iData(iData% MOD)
{
}
C1097Int(T2 llData) :m_iData(llData% MOD) {
}
C1097Int operator+(const C1097Int& o)const
{
return C1097Int(((T2)m_iData + o.m_iData) % MOD);
}
C1097Int& operator+=(const C1097Int& o)
{
m_iData = ((T2)m_iData + o.m_iData) % MOD;
return *this;
}
C1097Int& operator-=(const C1097Int& o)
{
m_iData = ((T2)MOD + m_iData - o.m_iData) % MOD;
return *this;
}
C1097Int operator-(const C1097Int& o)
{
return C1097Int(((T2)MOD + m_iData - o.m_iData) % MOD);
}
C1097Int operator*(const C1097Int& o)const
{
return((T2)m_iData * o.m_iData) % MOD;
}
C1097Int& operator*=(const C1097Int& o)
{
m_iData = ((T2)m_iData * o.m_iData) % MOD;
return *this;
}
C1097Int operator/(const C1097Int& o)const
{
return *this * o.PowNegative1();
}
C1097Int& operator/=(const C1097Int& o)
{
*this /= o.PowNegative1();
return *this;
}
bool operator==(const C1097Int& o)const
{
return m_iData == o.m_iData;
}
bool operator<(const C1097Int& o)const
{
return m_iData < o.m_iData;
}
C1097Int pow(T2 n)const
{
C1097Int iRet = (T1)1, iCur = *this;
while (n)
{
if (n & 1)
{
iRet *= iCur;
}
iCur *= iCur;
n >>= 1;
}
return iRet;
}
C1097Int PowNegative1()const
{
return pow(MOD - 2);
}
T1 ToInt()const
{
return ((T2)m_iData + MOD) % MOD;
}
private:
T1 m_iData = 0;;
};
class CParentToNeiBo
{
public:
CParentToNeiBo(const vector<int>& parents)
{
m_vNeiBo.resize(parents.size());
for (int i = 0; i < parents.size(); i++)
{
if (-1 == parents[i])
{
m_root = i;
}
else
{
m_vNeiBo[parents[i]].emplace_back(i);
}
}
}
vector<vector<int>> m_vNeiBo;
int m_root = -1;
};
class CBFSLeve {
public:
static vector<int> Leve(const vector<vector<int>>& neiBo, vector<int> start) {
vector<int> leves(neiBo.size(), -1);
for (const auto& s : start) {
leves[s] = 0;
}
for (int i = 0; i < start.size(); i++) {
for (const auto& next : neiBo[start[i]]) {
if (-1 != leves[next]) { continue; }
leves[next] = leves[start[i]] + 1;
start.emplace_back(next);
}
}
return leves;
}
template<class NextFun>
static vector<int> Leve(int N, NextFun nextFun, vector<int> start) {
vector<int> leves(N, -1);
for (const auto& s : start) {
leves[s] = 0;
}
for (int i = 0; i < start.size(); i++) {
auto nexts = nextFun(start[i]);
for (const auto& next : nexts) {
if (-1 != leves[next]) { continue; }
leves[next] = leves[start[i]] + 1;
start.emplace_back(next);
}
}
return leves;
}
static vector<vector<int>> LeveNodes(const vector<int>& leves) {
const int iMaxLeve = *max_element(leves.begin(), leves.end());
vector<vector<int>> ret(iMaxLeve + 1);
for (int i = 0; i < leves.size(); i++) {
ret[leves[i]].emplace_back(i);
}
return ret;
};
static vector<int> LeveSort(const vector<int>& leves) {
const int iMaxLeve = *max_element(leves.begin(), leves.end());
vector<vector<int>> leveNodes(iMaxLeve + 1);
for (int i = 0; i < leves.size(); i++) {
leveNodes[leves[i]].emplace_back(i);
}
vector<int> ret;
for (const auto& v : leveNodes) {
ret.insert(ret.end(), v.begin(), v.end());
}
return ret;
};
};
class CParents
{
public:
CParents(vector<int>& vParent, long long iMaxDepth)
{
int iBitNum = 0;
for (; iMaxDepth; iBitNum++) {
const auto mask = 1LL << iBitNum;
if (mask & iMaxDepth) { iMaxDepth = iMaxDepth ^ mask; }
}
const int n = vParent.size();
m_vParents.assign(iBitNum + 1, vector<int>(n, -1));
m_vParents[0] = vParent;
//树上倍增
for (int i = 1; i < m_vParents.size(); i++)
{
for (int j = 0; j < n; j++)
{
const int iPre = m_vParents[i - 1][j];
if (-1 != iPre)
{
m_vParents[i][j] = m_vParents[i - 1][iPre];
}
}
}
}
int GetParent(int iNode, int iDepth)const
{
int iParent = iNode;
for (int iBit = 0; iBit < m_vParents.size(); iBit++)
{
if (-1 == iParent)
{
return iParent;
}
if (iDepth & (1 << iBit))
{
iParent = m_vParents[iBit][iParent];
}
}
return iParent;
}
inline int GetBitCnt()const { return m_vParents.size(); };
inline const int& GetPow2Parent(int iNode, int n)const {
return m_vParents[n][iNode];
}
//在leftNodeExclude的1到rightLeve级祖先中查找符合pr的最近祖先
template<class _Pr>
int FindFirst(int leftNodeExclude, int rightLeve, _Pr pr) {
for (int iBit = GetBitCnt() - 1; iBit >= 0; iBit--) {
const int iMask = 1 << iBit;
if (!(iMask & rightLeve)) { continue; }
if (pr(m_vParents[iBit][leftNodeExclude])) {
return BFindFirst(leftNodeExclude, iBit, pr);
}
leftNodeExclude = m_vParents[iBit][leftNodeExclude];
}
return -1;
}
protected:
//在leftNodeExclude的1到2^pow^级祖先中查找符合pr的最近祖先
template<class _Pr>
int BFindFirst(int leftNodeExclude, int pow, _Pr pr) {
while (pow > 0) {
const int& mid = m_vParents[pow - 1][leftNodeExclude];
if (pr(mid)) {
}
else {
leftNodeExclude = mid;
}
pow--;
}
return m_vParents[0][leftNodeExclude];
}
vector<vector<int>> m_vParents;
};
class CNeiBo
{
public:
static vector<vector<int>> Two(int n, const vector<pair<int, int>>& edges, bool bDirect, int iBase = 0)
{
vector<vector<int>> vNeiBo(n);
for (const auto& [i1, i2] : edges)
{
vNeiBo[i1 - iBase].emplace_back(i2 - iBase);
if (!bDirect)
{
vNeiBo[i2 - iBase].emplace_back(i1 - iBase);
}
}
return vNeiBo;
}
static vector<vector<int>> Two(int n, const vector<vector<int>>& edges, bool bDirect, int iBase = 0)
{
vector<vector<int>> vNeiBo(n);
for (const auto& v : edges)
{
vNeiBo[v[0] - iBase].emplace_back(v[1] - iBase);
if (!bDirect)
{
vNeiBo[v[1] - iBase].emplace_back(v[0] - iBase);
}
}
return vNeiBo;
}
static vector<vector<std::pair<int, int>>> Three(int n, vector<vector<int>>& edges, bool bDirect, int iBase = 0)
{
vector<vector<std::pair<int, int>>> vNeiBo(n);
for (const auto& v : edges)
{
vNeiBo[v[0] - iBase].emplace_back(v[1] - iBase, v[2]);
if (!bDirect)
{
vNeiBo[v[1] - iBase].emplace_back(v[0] - iBase, v[2]);
}
}
return vNeiBo;
}
static vector<vector<std::pair<int, int>>> Three(int n, const vector<tuple<int, int, int>>& edges, bool bDirect, int iBase = 0)
{
vector<vector<std::pair<int, int>>> vNeiBo(n);
for (const auto& [u, v, w] : edges)
{
vNeiBo[u - iBase].emplace_back(v - iBase, w);
if (!bDirect)
{
vNeiBo[v - iBase].emplace_back(u - iBase, w);
}
}
return vNeiBo;
}
static vector<vector<int>> Mat(vector<vector<int>>& neiBoMat)
{
vector<vector<int>> neiBo(neiBoMat.size());
for (int i = 0; i < neiBoMat.size(); i++)
{
for (int j = i + 1; j < neiBoMat.size(); j++)
{
if (neiBoMat[i][j])
{
neiBo[i].emplace_back(j);
neiBo[j].emplace_back(i);
}
}
}
return neiBo;
}
};
class Solution {
public:
vector<float> Ans(int N, const vector<tuple<int, int, int>>& edge) {
m_iN = N;
m_neiBo = CNeiBo::Three(N, edge, false, 1);
m_vk.assign(N, -100); m_vb.assign(N, 0);
for (int i = 0; i < N; i++)
{
if (-100 != m_vk[i]) { continue; }
BFS(i);
}
if (!Check(m_vb, edge)) { return {}; }
return m_vb;
}
void BFS(int root) {
vector<int> que; vector<bool> vis(m_iN);
vector<float> bs;
que.emplace_back(root); vis[root] = true; m_vk[root] = 1;
float x = 2e10;
for (int i = 0; i < que.size(); i++) {
const auto cur = que[i];
for (const auto& [next, w] : m_neiBo[cur]) {
const auto k1 = -m_vk[cur];
const auto b1 = w - m_vb[cur];
if (-100 == m_vk[next]) {
m_vk[next] = k1; m_vb[next] = b1;
}
else if ((x > 1e10) && (m_vk[next] != k1)) {
x = (m_vb[next] - b1) / (k1 - m_vk[next]);
}
if (!vis[next]) {
vis[next] = true; que.emplace_back(next);
}
}
}
if (x > 1e10)
{
for (const auto& cur : que) {
if (1 == m_vk[cur]) {
bs.emplace_back(-m_vb[cur]);
}
else if (-1 == m_vk[cur]) {
bs.emplace_back(m_vb[cur]);
}
}
if (bs.size()) {
nth_element(bs.begin(), bs.begin() + bs.size() / 2, bs.end());
x = bs[bs.size() / 2];
}
}
for (const auto& cur : que) {
m_vb[cur] += m_vk[cur] * x;
}
}
static bool Check(vector<float>& res, vector<tuple<int, int, int>> edge) {
for (auto [u, v, w] : edge) {
u--, v--;
if (abs(res[u] + res[v] - w) > 0.00001) { return false; }
}
return true;
}
int m_iN;
vector<int> m_vk;
vector<float> m_vb;
vector<vector<std::pair<int, int>>> m_neiBo;
};
int main() {
#ifdef _DEBUG
freopen("a.in", "r", stdin);
#endif // DEBUG
ios::sync_with_stdio(0); cin.tie(nullptr);
int n;
cin >> n;
auto edge = Read<tuple<int, int, int>>();
#ifdef _DEBUG
printf("N=%d",n);
//cout << ",s=" << s;
Out(edge, ",edge=");
//Out(ws, ",hs=");
//Out(que, ",que=");
/*Out(que, "que=");*/
#endif // DEBUG
auto res = Solution().Ans(n,edge);
cout << (res.size()?"YES":"NO" )<< '\n';
for (const auto& f : res) {
cout << f << " ";
}
return 0;
}
单元测试
cpp
void Check(vector<float>& res ,float sum, const vector<tuple<int, int, int>>& edge) {
float actsum = 0;
for (const auto& f : res) {
actsum += abs(f);
}
Assert::IsTrue(abs(actsum - sum) < 0.00001);
Assert::IsTrue(Solution::Check(res, edge));
}
int N;
vector<tuple<int, int, int>> edge;
TEST_METHOD(TestMethod1)
{
N = 4, edge = { {1,2,1},{2,3,2},{1,3,2},{3,4,1} };
auto res = Solution().Ans(N, edge);
Check(res, 3, edge);
}
TEST_METHOD(TestMethod2)
{
N = 2, edge = { {1,2,1} };
auto res = Solution().Ans(N, edge);
Check(res, 1, edge);
}
TEST_METHOD(TestMethod3)
{
N = 3, edge = { {1,2,2},{2,3,2} };
auto res = Solution().Ans(N, edge);
Check(res, 2, edge);
}
TEST_METHOD(TestMethod4)
{
N = 3, edge = { {1,2,2},{2,2,1},{2,1,1},{1,2,2} };
auto res = Solution().Ans(N, edge);
AssertEx(0u, res.size());
}
扩展阅读
我想对大家说的话 |
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工作中遇到的问题,可以按类别查阅鄙人的算法文章,请点击《算法与数据汇总》。 |
学习算法:按章节学习《喜缺全书算法册》,大量的题目和测试用例,打包下载。重视操作 |
有效学习:明确的目标 及时的反馈 拉伸区(难度合适) 专注 |
闻缺陷则喜(喜缺)是一个美好的愿望,早发现问题,早修改问题,给老板节约钱。 |
子墨子言之:事无终始,无务多业。也就是我们常说的专业的人做专业的事。 |
如果程序是一条龙,那算法就是他的是睛 |
失败+反思=成功 成功+反思=成功 |
视频课程
先学简单的课程,请移步CSDN学院,听白银讲师(也就是鄙人)的讲解。
https://edu.csdn.net/course/detail/38771
如何你想快速形成战斗了,为老板分忧,请学习C#入职培训、C++入职培训等课程
https://edu.csdn.net/lecturer/6176
测试环境
操作系统:win7 开发环境: VS2019 C++17
或者 操作系统:win10 开发环境: VS2022 C++17
如无特殊说明,本算法用**C++**实现。