题意：求以1为根的最小树形图 没有输出字符串
思路：直接高朱刘算法 不懂的可以百度 学会了就是直接套模板的事情 其实就是不断消圈而已 不构成圈就有解 无法从根到达其他点就无解
#include <cstdio>
#include <cstring>
#include <cmath>
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;
}