要想去除边，并且不改变连通性，而且去除的值最大，相当于保留最小生成树。
注意这题连通块有若干个，所以运行Kruskal相当于形成若干个最小生成树。
如果是prim只能事先处理好各个连通块，然后在连通块内部单独用prim

题目链接

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;
const int N = 110;
const int M = 10010;struct Edge {int a, b, w;bool operator< (const Edge &t) const {return w < t.w;}
};Edge e[M];
int p[N];
int n, m;int find(int x) {if (p[x] != x) p[x] = find(p[x]);return p[x];
}
int kruskal() {for (int i = 1; i <= n; i ++ ) p[i] = i;sort(e, e + m);int res = 0;for (int i = 0; i < m; i ++ ) {int a = find(e[i].a);int b = find(e[i].b);if (a != b) {p[a] = b;}else {res += e[i].w;}}return res;
}
int main() {scanf("%d%d", &n, &m);int a, b, w;for (int i = 0; i < m; i ++ ) {scanf("%d%d%d", &a, &b, &w);e[i] = {a, b, w};}cout << kruskal() << endl;
}