AcWing 858. Prim算法求最小生成树

//y总做法
#include <bits/stdc++.h>
using namespace std;
const int N = 510,INF=0x3f3f3f3f;
int n, m;
int g[N][N];
bool st[N];
int dist[N];
int prim() {
    memset(dist, 0x3f, sizeof dist);
    int res = 0;
    dist[n]=0;
    for (int i = 0; i < n; i++) {
        int t = -1;
        for (int j = 1; j <= n; j++)
            if (!st[j] && (t == -1 || dist[t] > dist[j])) t = j;
        if (i && dist[t] == INF) return INF;
        if (i) res += dist[t];
        for (int j = 1; j <= n; j++) dist[j] = min(dist[j], g[t][j]);
        st[t] = true;
    }
    return res;
}
int main() {
    scanf("%d%d", &n, &m);
    memset(g, 0x3f, sizeof g);
    while (m--) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        g[a][b] = g[b][a] = min(g[a][b], c);
    }
    int t = prim();
    if (t == INF) printf("impossible");
    else printf("%d", t);
    return 0;
}
/*
//算法竞赛进阶指南
#include <bits/stdc++.h>
using namespace std;
const int N=510,INF=0x3f3f3f3f;
int n,m;
int g[N][N],ans;
//dist表示点到集合的距离
int dist[N];
bool st[N];
void prim(){
    memset(dist,0x3f,sizeof dist);
    dist[1]=0;
    for(int i=1;i<n;i++){
        int t=-1;
        for(int j=1;j<=n;j++)
            if(!st[j]&&(t==-1||dist[t]>dist[j]))
                t=j;
        st[t]=true;
        for(int j=1;j<=n;j++) 
            if(!st[j]) dist[j]=min(dist[j],g[t][j]);

    }

}
int main(){
    scanf("%d%d",&n,&m);
    memset(g,0x3f,sizeof g);
    while(m--){
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        g[a][b]=g[b][a]=min(g[a][b],c);
    }
    prim();
    for(int i=2;i<=n;i++) ans+=dist[i];
    if(ans>INF/2) printf("impossible");
    else printf("%d",ans);
    return 0;
}
*/