这题就是最小生成树Prim，d[]代表的是整个连接集合和未连接点的最小距离。

#include 
    
     int map[105][105];int vis[105], d[105];const int INF = 0x3f3f3f3f;int n,ans;void Prim(){
        d[0] = 0;    ans = 0;    for (int i = 0; i < n;i++) {
            int end = -1;        for (int j = 0; j < n;j++) {
                if (!vis[j]&&(end==-1||d[j]
     
      map[end][j])                d[j] = map[end][j];        }    }}int main(){
        while (scanf("%d",&n)!=EOF) {
            for (int i = 0; i < 105;i++) {
                vis[i] = 0;            d[i] = INF;        }        for (int i = 0; i < n;i++) {
                for (int j = 0; j < n;j++) {
                    scanf("%d",&map[i][j]);            }        }        Prim();        printf("%d\n", ans);    }    return 0;}