题目链接：http://poj.org/problem?id=3164

       题意是有n个点m条边，然后输入n个点的坐标，然后输入m个u,v表示相连的两个点，权值为两点间距离，边为单向边，问能否将n个点连起来，且花费最少。

       其实就是一个有向图的最小生成树，叫做最小树形图，用朱刘算法写，这道题直接套模板就好了，但是tmd有坑点，最后的结果要输出%.2f，不能是%.2lf，就因为这个卡了我一下午+一晚上，让我一直以为是板子哪里有问题...

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 = 1010*1010;
struct Node{
  double x,y;
}p[MAXN];
struct Edge{
    int u, v;
    double cost;
}edge[MAXM];
int pre[MAXN], id[MAXN], vis[MAXN];
double in[MAXN];
int n,m,root;

double dist(Node a,Node b){
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

double solve()
{
    double 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()
{
  while(~scanf("%d%d",&n,&m)){
    for(int i=0;i<n;i++){
      scanf("%lf%lf",&p[i].x,&p[i].y);
    }
    for(int i=0;i<m;i++){
			scanf("%d%d",&edge[i].u,&edge[i].v);
			edge[i].u --; edge[i].v --;
			if(edge[i].u != edge[i].v)
			edge[i].cost = dist(p[edge[i].u], p[edge[i].v]);
			else edge[i].cost = INF;
		}
		root = 0;
		double ans = solve();
		if(ans == -1) printf("poor snoopy\n");
		else printf("%.2f\n", ans);
  }
  return 0;
}