POJ 3164 Command Network 最小樹形圖-朱劉算法裸題

題目來源：POJ 3164 Command Network

題意：求以1為根的最小樹形圖 沒有輸出字符串

思路：直接高朱劉算法 不懂的可以百度 學會了就是直接套模板的事情 其實就是不斷消圈而已 不構成圈就有解 無法從根到達其他點就無解

#include 
#include 
#include 
const int maxn = 110;
const int maxm = 50010;
const double INF = 999999999;
struct edge
{
	int u, v;
	double w;
	edge(){}
	edge(int u, int v, double w) : u(u), v(v), w(w){}
}e[maxm];

struct Point
{
	double x, y;
}p[maxn];

int pre[maxn], vis[maxn], no[maxn];
double in[maxn];
double dis(Point a, Point b)  
{  
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));  
}  
double MST(int n, int m, int rt)
{
	double ans = 0;
	while(1)
	{
		for(int i = 1; i <= n; i++)
			in[i] = INF;
		for(int i = 1; i <= m; i++)
		{
			int u = e[i].u;
			int v = e[i].v;
			if(u != v && in[v] > e[i].w)
			{
				pre[v] = u;
				in[v] = e[i].w;
			}
		}
		for(int i = 1; i <= n; i++)
		{
			if(i == rt)
				continue;
			if(in[i] == INF)
				return -1;
		}
		memset(no, -1, sizeof(no));
		memset(vis, -1, sizeof(vis));
		in[rt] = 0;
		int cnt = 0;
		for(int i = 1; i <= n; i++)
		{
			ans += in[i];
			int v = i;
			while(v != rt && vis[v] != i && no[v] == -1)
			{
				vis[v] = i;
				v = pre[v];
			}
			if(v != rt && no[v] == -1)
			{
				for(int u = pre[v]; u != v; u = pre[u])
					no[u] = cnt+1;
				no[v] = cnt+1;
				cnt++;
			}
		}
		if(!cnt)
			return ans;
		for(int i = 1; i <= n; i++)
			if(no[i] == -1)
				no[i] = ++cnt;
		for(int i = 1; i <= m; i++)
		{
			int u = e[i].u;
			int v = e[i].v;
			e[i].u = no[u];
			e[i].v = no[v];
			if(no[u] != no[v])
			{
				e[i].w -= in[v];
			}
		}
		n = cnt;
		rt = no[rt];
	}
	return ans;
}
int main()
{
	int n, m;
	while(scanf("%d %d", &n, &m) != EOF)
	{
		for(int i = 1 ; i <= n; i++)
			scanf("%lf %lf", &p[i].x, &p[i].y);
		int add = 1;
		for(int i = 1; i <= m; i++)
		{
			scanf("%d %d", &e[add].u, &e[add].v);
			if(e[add].u == e[add].v)
				continue;
			e[add].w = dis(p[e[add].u], p[e[add].v]);
			add++;
		}
		double ans = MST(n, add-1, 1);
		if(ans < 0)
			puts("poor snoopy");
		else
			printf("%.2f\n", ans);
	}
	return 0;
}