poj-1751

Highways

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13300		Accepted: 3846		Special Judge

Description

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor system of public highways. The Flatopian government is aware of this problem and has already constructed a number of highways connecting some of the most important towns. However, there are still some towns that you can't reach via a highway. It is necessary to build more highways so that it will be possible to drive between any pair of towns without leaving the highway system. 

Flatopian towns are numbered from 1 to N and town i has a position given by the Cartesian coordinates (xi, yi). Each highway connects exaclty two towns. All highways (both the original ones and the ones that are to be built) follow straight lines, and thus their length is equal to Cartesian distance between towns. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways. 

The Flatopian government wants to minimize the cost of building new highways. However, they want to guarantee that every town is highway-reachable from every other town. Since Flatopia is so flat, the cost of a highway is always proportional to its length. Thus, the least expensive highway system will be the one that minimizes the total highways length. 

Input

The input consists of two parts. The first part describes all towns in the country, and the second part describes all of the highways that have already been built. 

The first line of the input file contains a single integer N (1 <= N <= 750), representing the number of towns. The next N lines each contain two integers, xi and yi separated by a space. These values give the coordinates of i ^th town (for i from 1 to N). Coordinates will have an absolute value no greater than 10000. Every town has a unique location. 

The next line contains a single integer M (0 <= M <= 1000), representing the number of existing highways. The next M lines each contain a pair of integers separated by a space. These two integers give a pair of town numbers which are already connected by a highway. Each pair of towns is connected by at most one highway. 

Output

Write to the output a single line for each new highway that should be built in order to connect all towns with minimal possible total length of new highways. Each highway should be presented by printing town numbers that this highway connects, separated by a space. 

If no new highways need to be built (all towns are already connected), then the output file should be created but it should be empty. 

Sample Input

9
1 5
0 0 
3 2
4 5
5 1
0 4
5 2
1 2
5 3
3
1 3
9 7
1 2

Sample Output

1 6
3 7
4 9
5 7
8 3

Source

Northeastern Europe 1999

用的是 kruskal 算法  对每一个点进行标号 ,第一次学会把坐标转换成每个点具有的标号
WA了一个下午，终于让我AC了，废话不多说，先翻译
9 9代表9个村庄1-9 
1 5 第1个村庄的坐标(1,5) 
0 0 第2个村长的坐标(0,0) 下面的一次类推 
3 2
4 5
5 1
0 4
5 2
1 2 (让你求剩下的村庄 输出哪个村庄和哪个村庄相连 最小生成树) 
5 3
3 3代表已经建好了3条路 
1 3 1、3相连 
9 7 9、7相连 
1 2 1、2相连 
这题太坑了，第一次尝试用这个算法，就碰到一个坎
一定要注意n个点至少需要n-1条边，但是这道题目已经给你出来几条路了
他们有可能给出的路可以围成一个环，所以这里的路可能大于n-1条 写了一下午
多亏学长 指点 代码G++才能够(⊙o⊙)哦 

#include<cstdio>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int pre[1000];
int n;//村庄的数目
int x[1000],y[1000];
struct node
{
	int a;
	int b;
	double len;
}arr[600000];
bool cmp(node c,node d)
{
	return c.len<d.len;
}

double dis(int c,int d)
{
	return sqrt((x[c]-x[d])*(x[c]-x[d])*1.0+(y[c]-y[d])*(y[c]-y[d])*1.0);
}
void count()
{
	for(int i=1;i<=n;++i)
		pre[i]=i;
} 

int find(int u)
{
	int i=u;
	while(pre[u]!=u)
		u=pre[u];
	int j;
	while(pre[i]!=u)
	{
		j=pre[i];
		pre[i]=u;
		i=j;
	}
	return u;
}

int main()
{
	scanf("%d",&n);
	int i;
	int sum=0;
	count();//初始化数据 
	for(i=0;i<n;++i)
	{
		scanf("%d%d",&x[i],&y[i]);
	}
	int m;
	scanf("%d",&m);
	for(i=0;i<m;++i)
	{
		int a,b;
		scanf("%d%d",&a,&b);//建立起来联系
		int fa=find(a);
		int fb=find(b);
		if(fa!=fb)
		{
			pre[fa]=fb;
			++sum;
		}	 
	}
	int j;
	int ant=0;
	for(i=0;i<n-1;++i)
	{
		for(j=i+1;j<n;++j)
		{
			arr[ant].a=i+1;arr[ant].b=j+1;
			arr[ant].len=dis(i,j);
			++ant;
		}
	}
	sort(arr,arr+ant,cmp);
	for(i=0;i<ant;++i)
	{
		int v=find(arr[i].a);
		int w=find(arr[i].b);
		if(v!=w)
		{
			pre[v]=w;
			++sum;
			printf("%d %d\n",arr[i].a,arr[i].b);
			if(sum==n-1)
				break;
		}
	}
	
}