Aizu 2224 Save your cats【最大生成树】

Description

Nicholas Y. Alford was a cat lover. He had a garden in a village and kept many cats in his garden. The cats were so cute that people in the village also loved them.

One day, an evil witch visited the village. She envied the cats for being loved by everyone. She drove magical piles in his garden and enclosed the cats with magical fences running between the piles. She said “Your cats are shut away in the fences until they become ugly old cats.” like a curse and went away.

Nicholas tried to break the fences with a hummer, but the fences are impregnable against his effort. He went to a church and asked a priest help. The priest looked for how to destroy the magical fences in books and found they could be destroyed by holy water. The Required amount of the holy water to destroy a fence was proportional to the length of the fence. The holy water was, however, fairly expensive. So he decided to buy exactly the minimum amount of the holy water required to save all his cats. How much holy water would be required?

Input

The input has the following format:

N M
x₁ y₁
.
.
.
x_N y_N
p₁ q₁
.
.
.
p_M q_M

The first line of the input contains two integers N (2 ≤ N ≤ 10000) and M (1 ≤ M). N indicates the number of magical piles and Mindicates the number of magical fences. The following N lines describe the coordinates of the piles. Each line contains two integers x_iand y_i (-10000 ≤ x_i, y_i ≤ 10000). The following M lines describe the both ends of the fences. Each line contains two integers p_j and q_j (1 ≤p_j, q_j ≤ N). It indicates a fence runs between the p_j-th pile and the q_j-th pile.

You can assume the following:

• No Piles have the same coordinates.
• A pile doesn’t lie on the middle of fence.
• No Fences cross each other.
• There is at least one cat in each enclosed area.
• It is impossible to destroy a fence partially.
• A unit of holy water is required to destroy a unit length of magical fence.

Output

Output a line containing the minimum amount of the holy water required to save all his cats. Your program may output an arbitrary number of digits after the decimal point. However, the absolute error should be 0.001 or less.

Sample Input 1

3 3
0 0
3 0
0 4
1 2
2 3
3 1

Output for the Sample Input 1

3.000

Sample Input 2

4 3
0 0
-100 0
100 0
0 100
1 2
1 3
1 4

Output for the Sample Input 2

0.000

Sample Input 3

6 7
2 0
6 0
8 2
6 3
0 5
1 7
1 2
2 3
3 4
4 1
5 1
5 4
5 6

Output for the Sample Input 3

7.236

Sample Input 4

6 6
0 0
0 1
1 0
30 0
0 40
30 40
1 2
2 3
3 1
4 5
5 6
6 4

Output for the Sample Input 4

31.000

有N个树桩的坐标，然后下面给出其中的M条边，边围成环时，要去掉每个环中的最小的一条边，把问题转化为求最大生成树，并记录边权值的和，所有边的和减去生成树的和

边的数组要开到大于10000*10000/2.不然Runtime Error

代码:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define MAXN 10010
using namespace std;

int per[MAXN];
struct node
{
	int a;
	int b;
	double edge;
 };
 node es[MAXN*MAXN];//最多能形成10000*10000/2 条边
 bool cmp(node x,node y)
 {
 	return x.edge>y.edge;
 }
 int find(int x)
 {
 	if(per[x]==x)
 		return x;
 	return per[x]=find(per[x]);
 }
 double ans;
 double kruskal(int n,int m)
 {
 	sort(es,es+m,cmp);
 	ans=0;
 	int cnt=0;
 	for(int i=0;i<m&&cnt<n-1;i++)
 	{
 		int fx=find(es[i].a);
 		int fy=find(es[i].b);
 		if(fx!=fy)
 		{
 		    per[fx]=fy;
 			ans+=es[i].edge;
 			cnt++;
		 }
	 }
	 return ans;
 }
 int main()
 {
 	int n,m;
 	double pointx[10000+10],pointy[10000+10];
 	while(~scanf("%d%d",&n,&m))
 	{
 		for(int i=1;i<=n;i++)
 		{
 			per[i]=i;
		 }
 		double sum=0.0;
 		for(int i=1;i<=n;i++)
 			scanf("%lf%lf",&pointx[i],&pointy[i]);
 		for(int i=0;i<m;i++)
 		{
 			scanf("%d%d",&es[i].a,&es[i].b);
 			es[i].edge=sqrt((pointx[es[i].a]-pointx[es[i].b])*(pointx[es[i].a]-pointx[es[i].b])+(pointy[es[i].a]-pointy[es[i].b])*(pointy[es[i].a]-pointy[es[i].b]));
 			sum+=es[i].edge;
		 }
		 printf("%.4lf\n",sum-kruskal(n,m));
	 }
	 return 0;
 }