Aizu 2224 (最大生成树，并查集)

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

x1y1

.

.

.

xNyN

p1q1

.

.

.

pMqM

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 xi andyi (-10000 ≤ xi, yi ≤ 10000). The following M lines describe the both ends of the fences. Each line contains two integers pj and qj (1 ≤ pj,qj ≤ N). It indicates a fence runs between the pj-th pile and the qj-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

题意：这道题呢就是给出n个点，m条线，m条线可能围出封闭的多边形，让求出破坏一条短的边，求破坏的最小长度

这题我刚看，没思路，也没有做过，第一次用并查集做最大生成树，我也是参考着别人的代码，首先我们用s去计算总的长度，然后用sum去求最大生成树的总长度，得到的结果就是我们所要求的结果，我们该怎么求最大的生成树呢，当然用prim算法是不行的，我们要用并查集，我们先让所有的边从大到小的顺序排序一下，

把并查集f[i]赋值一下，当加入某条边的时候，若f[x]==x,则返回 x,否则的话就并归 f[x]=Find(f[x]),return f[x]，相同的话，会构成一个封闭区间

还有啊，数组呢，要开大点因为m的范围不定，开小了，就出现time error

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
const int N=1000100;
struct node
{
int l,r;
double cost;
}p
;
double  x
,y
;
int f
;
int Find(int x)
{
if(x==f[x])
return x;
else
{
f[x]=Find(f[x]);
return f[x];
}
}
int merg(int a,int b)
{
int t1=Find(a);
int t2=Find(b);
if(t1!=t2)
{
f[t2]=t1;
return 0;
}
return 1;
}
int cmp(node a,node b)
{
return a.cost>b.cost;//边是从大到小排的
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int i,j;
for(i=1;i<=n;i++)
{
scanf("%lf%lf",&x[i],&y[i]);
}
double s=0;
for(i=0;i<m;i++)
{
scanf("%d%d",&p[i].l,&p[i].r);
int a=p[i].l,b=p[i].r;
p[i].cost=sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));
s+=p[i].cost;
}
for(i=1;i<=n;i++)
{
f[i]=i;
}
sort(p,p+m,cmp);
double sum=0;
for(i=0;i<m;i++)
{
if(!merg(p[i].l,p[i].r))
sum+=p[i].cost;
}
printf("%.3lf\n",s-sum);
}
return 0;
}