最小生成树模板

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define LL long long
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 900000r
#define N 1000
int n,m;
int ans;
int mp[N][N];
int vis[N];
struct edge
{
    int to;//to为通向的边
    int v;//为权值
    friend bool operator<(const edge& a,const edge& b)
    {
        return a.v>b.v;
    }
}now;

void prim(int s)
{
    vis[s]=1;
    priority_queue<edge>q;
    while(!q.empty())q.pop();

    rep(i,1,n)
    {
        rep(j,1,n)
        if(!vis[j])
        {
            now.to=j;
            now.v=mp[s][j];
            q.push(now);
        }
        while(!q.empty()&&vis[q.top().to])
            q.pop();
        if(q.empty())break;
        now=q.top();q.pop();
        s=now.to;
        ans+=now.v;
        vis[s]=1;
    }
}

int main()
{
    CLR(vis,0);
    RII(n,m);
    rep(i,1,n)
    rep(j,1,n)
    mp[i][j]=inf;
    while(m--)
    {
        int a,b,c;
        RIII(a,b,c);
        a++;b++;
        mp[a][b]=mp[b][a]=c;
    }
    prim(1);
    cout<<ans<<endl;
    return 0;
}