题目

Prim （生成一颗包含起点的最小生成树，所以要多次调用）

#include <bits/stdc++.h>using namespace std;const int N = 510;
const int inf = 0x3f3f3f3f;int n, m;
int g[N][N], dis[N];
bool p[N], vis[N];int prim (int u)
{memset(dis, 0x3f, sizeof dis); dis[u] = 0;int sum = 0;for(int i = 0 ; i < n ; i ++ ){int t = -1;for(int j = 1 ; j <= n ; j ++ )if(!p[j] && (t == -1 || dis[t] > dis[j]))t = j;if(i && dis[t] == inf) return sum;p[t] = 1;if(i) sum += dis[t]; // 第一个点为根节点没有边权vis[t] = 1;for(int j = 1 ; j <= n ; j ++ )if(!p[j] && dis[j] > g[t][j]) dis[j] = g[t][j];}return sum;
}int main ()
{int ans = 0;cin >> n >> m;for(int i = 1 ; i <= n ; i ++ )for(int j = 1 ; j <= n ; j ++ )if(i == j) g[i][j] = 0;else g[i][j] = inf;for(int i = 1 ; i <= m ; i ++ ){int a, b, c;cin >> a >> b >> c;g[a][b] = g[b][a] = min(g[a][b], c);ans += min(g[a][b], c);}int t = 0;for(int i = 1 ; i <= n ; i ++ )if(!vis[i]) t += prim(i);cout << ans - t << endl;return 0;
}

Kruskal （如果有多颗生成树，生成最小生成森林）

#include <bits/stdc++.h>
using namespace std;
const int N = 110;
const int M = 210;
struct edge{int a;int b;int c;bool operator < (const edge& v){return c < v.c;}
} e[M];
int p[N];
int n, idx, m;
int find(int x)
{if(p[x] != x) p[x] = find(p[x]);return p[x];
}
int kruskal()
{int retv = 0;for(int i = 1; i <= n; i++)p[i] = i;sort(e+1,e+m+1);for(int i = 1; i <= m; i++){int a = e[i].a, b = e[i].b, c = e[i].c;a = find(a), b = find(b);if(a != b){p[a] = b;retv += c;}}return retv;
}
int main()
{cin >> n >> m;int sum = 0;for(int i = 1; i <= m; i++){int a, b, c;cin >> a >> b >> c;e[++idx]  = {a, b, c};sum += c;}int t = kruskal();cout << sum - t;
}