acwing算法提高之图论--最小生成树的典型应用

1 介绍

本专题用来记录使用prim算法或kruskal算法求解的题目。

2 训练

题目1：1140最短网络

C++代码如下，

#include <iostream>
#include <cstring>using namespace std;const int N = 110, INF = 0x3f3f3f3f;
int g[N][N];
int d[N];
bool st[N];
int n, m;void prim() {memset(d, 0x3f, sizeof d);int res = 0;for (int i = 0; i < n; ++i) {int t = -1;for (int j = 1; j <= n; ++j) {if (!st[j] && (t == -1 || d[t] > d[j])) {t = j;}}st[t] = true;if (i) res += d[t];for (int j = 1; j <= n; ++j) {if (d[j] > g[t][j]) {d[j] = g[t][j];}}}cout << res << endl;return;
}int main() {cin >> n;for (int i = 1; i <= n; ++i) {for (int j = 1; j <= n; ++j) {cin >> g[i][j];}}prim();return 0;
}

题目2：1141局域网

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 110, M = 210;
int p[N];
int n, m;struct Edge {int a, b, w;bool operator< (const Edge &W) const {return w < W.w;}
}edges[M];int find(int x) {if (p[x] != x) p[x] = find(p[x]);return p[x];
}int main() {cin >> n >> m;int s = 0;for (int i = 0; i < m; ++i) {cin >> edges[i].a >> edges[i].b >> edges[i].w;s += edges[i].w;}for (int i = 1; i <= n; ++i) p[i] = i;sort(edges, edges + m);int res = 0, cnt = 0;for (int i = 0; i < m; ++i) {int a = edges[i].a, b = edges[i].b, w = edges[i].w;a = find(a);b = find(b);if (a != b) {p[a] = b;res += w;cnt++;}}cout << s - res << endl;return 0;
}

题目3：1142繁忙的都市

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 310, M = 8010;
int n, m;
int p[N];struct Edge {int a, b, w;bool operator< (const Edge &W) const {return w < W.w;}
}edges[M];int find(int x) {if (p[x] != x) p[x] = find(p[x]);return p[x];
}int main() {cin >> n >> m;for (int i = 1; i <= n; ++i) p[i] = i;for (int i = 0; i < m; ++i) {cin >> edges[i].a >> edges[i].b >> edges[i].w;}sort(edges, edges + m);int res = 0;int cnt = 0;for (int i = 0; i < m; ++i) {int a = edges[i].a, b = edges[i].b, w = edges[i].w;a = find(a);b = find(b);if (a != b) {p[a] = b;cnt += 1;res = w;}}cout << cnt << " " << res << endl;return 0;
}

题目4：