目录
- [1 介绍](#1 介绍)
- [2 训练](#2 训练)
1 介绍
本专题用来介绍使用最短路算法(spfa或dijkstra)与其它算法(dfs、二分、拓扑排序、动态规划等等)结合起来解题。
2 训练
题目1 :1135新年好
C++代码如下,
cpp
//版本1,使用vector来存储图,会有超时风险
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int N = 50010;
int n, m;
int dist[6][N];
bool st[N];
vector<vector<pair<int, int>>> g; //使用vector可能有超时分险
int source[6];
void spfa(int start, int dist[]) {
memset(dist, 0x3f, N * 4);
memset(st, 0, sizeof st);
dist[start] = 0;
queue<int> q;
q.push(start);
st[start] = true;
while (!q.empty()) {
auto t = q.front();
q.pop();
st[t] = false;
for (auto [b, w] : g[t]) {
if (dist[b] > dist[t] + w) {
dist[b] = dist[t] + w;
if (!st[b]) {
q.push(b);
st[b] = true;
}
}
}
}
return;
}
int dfs(int u, int start, int distance) {
if (u > 5) return distance;
int res = 0x3f3f3f3f;
for (int i = 1; i <= 5; ++i) {
if (!st[i]) {
int b = source[i];
st[i] = true;
res = min(res, dfs(u + 1, i, distance + dist[start][b]));
st[i] = false;
}
}
return res;
}
int main() {
cin >> n >> m;
g.resize(n + 10);
source[0] = 1;
for (int i = 1; i <= 5; ++i) cin >> source[i];
for (int i = 0; i < m; ++i) {
int a, b, c;
cin >> a >> b >> c;
g[a].emplace_back(b, c);
g[b].emplace_back(a, c);
}
for (int i = 0; i < 6; ++i) spfa(source[i], dist[i]);
memset(st, 0, sizeof st);
int res = dfs(1, 0, 0);
cout << res << endl;
return 0;
}
cpp
//版本2,使用h,e,ne,w数组来存储图
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 50010, M = 200010, INF = 0x3f3f3f3f;
int n, m;
int h[N], e[M], w[M], ne[M], idx;
int q[N], dist[6][N];
int source[6];
bool st[N];
void add(int a, int b, int c) {
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
void dijkstra(int start, int dist[]) {
memset(dist, 0x3f, N * 4);
dist[start] = 0;
memset(st, 0, sizeof st);
priority_queue<PII, vector<PII>, greater<PII>> heap;
heap.push({0, start});
while (heap.size()) {
auto t = heap.top();
heap.pop();
int ver = t.second;
if (st[ver]) continue;
st[ver] = true;
for (int i = h[ver]; ~i; i = ne[i]) {
int j = e[i];
if (dist[j] > dist[ver] + w[i]) {
dist[j] = dist[ver] + w[i];
heap.push({dist[j], j});
}
}
}
}
int dfs(int u, int start, int distance) {
if (u > 5) return distance;
int res = INF;
for (int i = 1; i <= 5; ++i) {
if (!st[i]) {
int next = source[i];
st[i] = true;
res = min(res, dfs(u + 1, i, distance + dist[start][next]));
st[i] = false;
}
}
return res;
}
int main() {
scanf("%d%d", &n, &m);
source[0] = 1;
for (int i = 1; i <= 5; ++i) scanf("%d", &source[i]);
memset(h, -1, sizeof h);
while (m--) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c), add(b, a, c);
}
for (int i = 0; i < 6; ++i) dijkstra(source[i], dist[i]);
memset(st, 0, sizeof st);
printf("%d\n", dfs(1, 0, 0));
return 0;
}
题目2 :340通信线路
C++代码如下,
cpp
#include <iostream>
#include <cstring>
#include <algorithm>
#include <deque>
using namespace std;
const int N = 1010, M = 20010;
int n, m, k;
int h[N], e[M], w[M], ne[M], idx;
int dist[N];
deque<int> q;
bool st[N];
void add(int a, int b, int c) {
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
bool check(int bound) {
memset(dist, 0x3f, sizeof dist);
memset(st, 0, sizeof st);
q.push_back(1);
dist[1] = 0;
while (q.size()) {
int t = q.front();
q.pop_front();
if (st[t]) continue;
st[t] = true;
for (int i = h[t]; ~i; i = ne[i]) {
int j = e[i], x = w[i] > bound;
if (dist[j] > dist[t] + x) {
dist[j] = dist[t] + x;
if (!x) q.push_front(j);
else q.push_back(j);
}
}
}
return dist[n] <= k;
}
int main() {
cin >> n >> m >> k;
memset(h, -1, sizeof h);
while (m--) {
int a, b, c;
cin >> a >> b >> c;
add(a, b, c), add(b, a, c);
}
int l = 0, r = 1e6 + 1;
while (l < r) {
int mid = l + r >> 1;
if (check(mid)) r = mid;
else l = mid + 1;
}
if (r == 1e6 + 1) cout << -1 << endl;
else cout << r << endl;
return 0;
}
题目3 :342道路与航线
C++代码如下,
cpp
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
const int N = 25010, M = 150010, INF = 0x3f3f3f3f;
int n, mr, mp, S;
int id[N];
int h[N], e[M], w[M], ne[M], idx;
int dist[N], din[N];
vector<int> block[N];
int bcnt;
bool st[N];
queue<int> q;
void add(int a, int b, int c) {
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
void dfs(int u, int bid) {
id[u] = bid, block[bid].push_back(u);
for (int i = h[u]; ~i; i = ne[i]) {
int j = e[i];
if (!id[j]) {
dfs(j, bid);
}
}
}
void dijkstra(int bid) {
priority_queue<PII, vector<PII>, greater<PII>> heap;
for (auto u : block[bid]) {
heap.push({dist[u], u});
}
while (heap.size()) {
auto t = heap.top();
heap.pop();
int ver = t.y, distance = t.x;
if (st[ver]) continue;
st[ver] = true;
for (int i = h[ver]; ~i; i = ne[i]) {
int j = e[i];
if (id[j] != id[ver] && --din[id[j]] == 0) q.push(id[j]);
if (dist[j] > dist[ver] + w[i]) {
dist[j] = dist[ver] + w[i];
if (id[j] == id[ver]) heap.push({dist[j], j});
}
}
}
}
void topsort() {
memset(dist, 0x3f, sizeof dist);
dist[S] = 0;
for (int i = 1; i <= bcnt; ++i) {
if (!din[i]) {
q.push(i);
}
}
while (q.size()) {
int t = q.front();
q.pop();
dijkstra(t);
}
}
int main() {
cin >> n >> mr >> mp >> S;
memset(h, -1, sizeof h);
while (mr--) {
int a, b, c;
cin >> a >> b >> c;
add(a, b, c), add(b, a, c);
}
for (int i = 1; i <= n; ++i) {
if (!id[i]) {
bcnt++;
dfs(i, bcnt);
}
}
while (mp--) {
int a, b, c;
cin >> a >> b >> c;
din[id[b]]++;
add(a, b, c);
}
topsort();
for (int i = 1; i <= n; ++i) {
if (dist[i] > INF / 2) cout << "NO PATH" << endl;
else cout << dist[i] << endl;
}
return 0;
}
题目4 :341最优贸易
C++代码如下,
cpp
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010, M = 2000010;
int n, m;
int w[N];
int hs[N], ht[N], e[M], ne[M], idx;
int dmin[N], dmax[N];
int q[N];
bool st[N];
void add(int h[], int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
void spfa(int h[], int dist[], int type) {
int hh = 0, tt = 1;
if (type == 0) {
memset(dist, 0x3f, sizeof dmin);
dist[1] = w[1];
q[0] = 1;
} else {
memset(dist, -0x3f, sizeof dmax);
dist[n] = w[n];
q[0] = n;
}
while (hh != tt) {
int t = q[hh++];
if (hh == N) hh = 0;
st[t] = false;
for (int i = h[t]; ~i; i = ne[i]) {
int j = e[i];
if (type == 0 && dist[j] > min(dist[t], w[j]) || type == 1 && dist[j] < max(dist[t], w[j])) {
if (type == 0) dist[j] = min(dist[t], w[j]);
else dist[j] = max(dist[t], w[j]);
if (!st[j]) {
q[tt++] = j;
if (tt == N) tt = 0;
st[j] = true;
}
}
}
}
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) scanf("%d", &w[i]);
memset(hs, -1, sizeof hs);
memset(ht, -1, sizeof ht);
while (m--) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(hs, a, b), add(ht, b, a);
if (c == 2) add(hs, b, a), add(ht, a, b);
}
spfa(hs, dmin, 0);
spfa(ht, dmax, 1);
int res = 0;
for (int i = 1; i <= n; ++i) res = max(res, dmax[i] - dmin[i]);
printf("%d\n", res);
return 0;
}