题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2124

       题意就是一个人xxx要给n-1个人打电话，但是有些人有其他人的联系方式，有些人没有，问xxx想要传达一条信息，能否让其余的n-1个人收到，能的话输出最小花费。

       就是一个有向图的最小生成树，用朱刘算法模板就好了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1010;
const int MAXM = 40005;
struct Edge{
    int u, v, cost;
}edge[MAXM];
int pre[MAXN], id[MAXN], vis[MAXN], in[MAXN];
int T,n,m,root;

int solve()
{
    int res = 0;
    while (1){
        for(int i=0;i<n;i++) in[i] = INF;
        for(int i=0;i<m;++i){
    			int u = edge[i].u, v = edge[i].v;
    			if(edge[i].cost < in[v] && u != v){
    				pre[v] = u;
						in[v] = edge[i].cost;
    			}
    		}
        for (int i = 0; i < n; i++){
            if (i != root && in[i] == INF) return -1;
        }
        int tn = 0;
        memset(id, -1, sizeof(id));
        memset(vis, -1, sizeof(vis));
        in[root] = 0;
        for(int i = 0; i < n; i++){
            res += in[i];
            int v = i;
            while(vis[v] != i && id[v] == -1 && v != root){
                vis[v] = i;
                v = pre[v];
            }
            if(v != root && id[v] == -1)
            {
                for(int u = pre[v]; u != v ; u = pre[u]) id[u] = tn;
                id[v] = tn++;
            }
        }
        if (tn == 0) break;
        for (int i = 0; i < n; i++){
            if(id[i] == -1) id[i] = tn++;
        }
        for (int i = 0; i < m; i++){
            int v = edge[i].v;
            edge[i].u = id[edge[i].u];
            edge[i].v = id[edge[i].v];
            if (edge[i].u != edge[i].v) edge[i].cost -= in[v];
        }
        n = tn;
        root = id[root];
    }
    return res;
}

int main()
{
  int Case = 1;
  scanf("%d",&T);
  while(T--){
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++){
			scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].cost);
		}
		root = 0;
		int ans = solve();
    printf("Case #%d: ",Case ++);
		if(ans == -1) printf("Possums!\n");
		else printf("%d\n", ans);
  }
  return 0;
}