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Kruskal算法要对边排序，然后打个并查集维护，但是实际上Prim有他好玩的地方，就把Dijkstra的到点的距离从dis[v]:dis[u]+w改成边dis[v]:w。

那肯定是Prim好写一点。Prim感觉复杂度是O((n+m)logn)，Kruskal是O(n+mlogm)。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 5005;

int n, m, s, t;
int dis[MAXN];
bool vis[MAXN];
vector<pair<int, int> > G[MAXN];

const int INF = 0x3f3f3f3f;

priority_queue<pair<int, int> >pq;
int Prim(int s) {
    for(int i = 1; i <= n; ++i)
        dis[i] = INF;
    int cnt = 0, sum = 0;
    dis[s] = 0;
    pq.push({-dis[s], s});
    while(!pq.empty()) {
        int u = pq.top().second;
        pq.pop();
        if(vis[u])
            continue;
        vis[u] = 1;
        ++cnt;
        sum += (dis[u]);
        for(auto e : G[u]) {
            int v = e.first, w = e.second;
            if(!vis[v] && w < dis[v]) {
                dis[v] = w;
                pq.push({-dis[v], v});
            }
        }
    }
    if(cnt == n)
        return sum;
    return -1;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; ++i) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        G[u].emplace_back(v, w);
        G[v].emplace_back(u, w);
    }
    int res = Prim(1);
    if(res == -1)
        puts("orz");
    else
        printf("%d
", res);
    return 0;
}