2023牛客OI赛前集训营-提高组（第三场）C.分糖果
文章目录
C-分糖果_2023牛客OI赛前集训营-提高组（第三场） (nowcoder.com)
题目大意
求前 i ( i ∈ [ 1 , n ] ) i(i\in[1, n]) i(i∈[1,n]) 个数分成 k k k 个连续的区间，每一个区间和的最大值最小是多少。
T T T 组数据
对于 30 p t s 30pts 30pts ， 1 ≤ n ≤ 100 , 1 ≤ k ≤ n 1 \le n \le 100 , 1\le k \le n 1≤n≤100,1≤k≤n
另外 20 p t s 20pts 20pts ， 1 ≤ n ≤ 1 0 4 , k = 1 1\le n \le 10^4 , k = 1 1≤n≤104,k=1
另外 50 p t s 50pts 50pts， 1 ≤ n ≤ 1 0 5 , 1 ≤ k ≤ n 1\le n \le 10^5 , 1\le k \le n 1≤n≤105,1≤k≤n
对于全部数据有 T ≤ 3 , ∣ a i ∣ ≤ 1 0 9 , 1 ≤ n ≤ 1 0 5 T \le 3 , |a_i| \le 10^9 , 1\le n \le 10^5 T≤3,∣ai∣≤109,1≤n≤105
做法
考试忘记加换行， 50 p t s → 0 50pts \to 0 50pts→0
对于 30 p t s 30pts 30pts
直接二分答案 + d p dp dp ， O ( n 2 ) O(n^2) O(n2) 判断能不能分成大于等于 k k k 块的和满足假设。
对于 20 p t s 20pts 20pts
直接前缀和乱搞
50 p t s 50pts 50pts 代码
cpp
#include <bits/stdc++.h>
#define fu(x , y , z) for(int x = y ; x <= z ; x ++)
#define LL long long
using namespace std;
const int N = 1e5 + 5 , M = 1e5 + 5;
const LL inf = 1e14 + 5;
int n , k;
LL a[M] , f[M] , s[M] , ans;
bool ck (LL x) {
int l = 1;
fu (i , 1 , n) f[i] = 0;
fu (i , 1 , n) {
if (i == 1) {
f[i] = (s[1] <= x);
}
else {
fu (j , 1 , i) {
if (s[i] - s[j - 1] <= x && (f[j - 1] || j == 1)) f[i] = max (f[i] , f[j - 1] + 1);
}
}
if (f[i] >= k) return 1;
}
return 0;
}
LL fans (LL l , LL r) {
if (l == r) {
return ck (l) ? l : inf;
}
LL mid = l + r >> 1;
if (ck (mid)) return min (fans (l , mid) , mid);
else return min (inf , fans (mid + 1 , r));
}
int main () {
// freopen ("candy2.in" , "r" , stdin);
int T;
scanf ("%d" , &T);
while (T --) {
scanf ("%d%d" , &n , &k);
fu (i , 1 , n) scanf ("%lld" , &a[i]) , s[i] = 0;
fu (i , 1 , n) s[i] = s[i - 1] + a[i];
if (k == 1) {
ans = inf;
fu (i , 1 , n) ans = min (ans , s[i]);
printf ("%lld\n" , ans);
continue;
}
printf ("%lld\n" , fans (-inf , inf));
}
}对于 100 p t s 100pts 100pts
权值线段树 + 离散化
我们发现上面的 d p dp dp 转移太慢了
s s s 数组表示前缀和 ，当前二分答案为 x x x
对于 i ∈ [ 1 , n ] i\in[1 , n] i∈[1,n] 我们只用找到 j ∈ [ 1 , i ] j\in[1 , i] j∈[1,i] 满足 s [ i ] − s [ j ] ≤ x s[i] - s[j] \le x s[i]−s[j]≤x 即 s [ i ] − x ≤ s [ j ] s[i] - x \le s[j] s[i]−x≤s[j] 时的最大值 + 1 +1 +1
即
f [ i ] = M A X j = 1 i f [ j ] + 1 ( s [ i ] − x ≤ s [ j ] ) f[i] = MAX_{j = 1}^i f[j] + 1(s[i] - x \le s[j]) f[i]=MAXj=1if[j]+1(s[i]−x≤s[j])
用权值线段树维护就好了。
初始化 f [ 0 ] = 0 f[0] = 0 f[0]=0
每次把 f [ i ] f[i] f[i] 插入权值线段树中
因为总和太大了，所以还要把 s [ i ] s[i] s[i] 和 s [ i ] − x s[i] - x s[i]−x 拿出来离散化（还有 0 0 0） ，动态开点。
cpp
#include <bits/stdc++.h>
#define fu(x , y , z) for(int x = y ; x <= z ; x ++)
#define LL long long
using namespace std;
const int N = 4e7 + 5 , M = 2e5 + 5;
const LL inf = 1e9 * 1e6;
const int inff = 1e9 + 5;
int n , k , cnt , mp[M];
LL a[M] , f[M] , s[M] , ans , ss[M << 1];
struct Tr {
int v , lp , rp;
} tr[N];
void glp (int p) {
if (!tr[p].lp) {
tr[p].lp = ++cnt;
tr[cnt].v = -inff;
}
}
void grp (int p) {
if (!tr[p].rp) {
tr[p].rp = ++cnt;
tr[cnt].v = -inff;
}
}
void change (int p , LL l , LL r , int x , int val) {
if (l == r) tr[p].v = max (tr[p].v , val);
else {
int mid = l + r >> 1;
if (x <= mid) {
glp (p);
change (tr[p].lp , l , mid , x , val);
}
else {
grp (p);
change (tr[p].rp , mid + 1 , r , x , val);
}
tr[p].v = max (tr[tr[p].lp].v , tr[tr[p].rp].v);
}
}
int query (int p , LL l , LL r , LL L , LL R) {
if (L <= l && R >= r) return tr[p].v;
else {
int mid = l + r >> 1 , ans1 = -inff , ans2 = -inff;
if (L <= mid && tr[p].lp)
ans1 = query (tr[p].lp , l , mid , L , R);
if (R > mid && tr[p].rp)
ans2 = query (tr[p].rp , mid + 1 , r , L , R);
return max (ans1 , ans2);
}
}
void cl (int p) { tr[p].lp = tr[p].rp = 0 , tr[p].v = -inff; }
void clear (int p , LL l , LL r) {
if (l == r) {
cl (p);
return;
}
int mid = l + r >> 1;
if (tr[p].lp)
clear (tr[p].lp , l , mid);
if (tr[p].rp)
clear (tr[p].rp , mid + 1 , r);
cl (p);
}
bool ck (LL x) {
int s1 = 1;
ss[1] = 0;
fu (i , 1 , n) {
ss[++s1] = s[i];
ss[++s1] = s[i] - x;
}
sort (ss + 1 , ss + s1 + 1);
int m = unique (ss + 1 , ss + s1 + 1) - ss - 1;
clear (1 , -M , M);
tr[0].v = -M;
cnt = 1;
int y = lower_bound(ss + 1 , ss + s1 + 1 , 0) - ss;
change (1 , -M , M , y , 0);
fu (i , 1 , n) {
y = lower_bound(ss + 1 , ss + s1 + 1 , s[i] - x) - ss;
f[i] = query (1 , -M , M , y , M) + 1;
y = lower_bound(ss + 1 , ss + s1 + 1 , s[i]) - ss;
change (1 , -M , M , y , f[i]);
if (f[i] >= k) return 1;
}
return 0;
}
LL fans (LL l , LL r) {
if (l == r) return ck (l) ? l : inf;
else {
LL mid = l + r >> 1;
if (ck (mid)) {
return min (fans (l , mid) , mid);
}
else {
return fans (mid + 1 , r);
}
}
}
int main () {
// freopen ("candy2.in" , "r" , stdin);
int T;
scanf ("%d" , &T);
while (T --) {
scanf ("%d%d" , &n , &k);
fu (i , 1 , n) scanf ("%lld" , &a[i]) , s[i] = 0;
fu (i , 1 , n) s[i] = s[i - 1] + a[i];
printf ("%lld\n" , fans (-inf , inf));
}
}