最小生成树-并查集-Kruskal-zoj-2048-special judge

Highways

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 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 ith 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 should be created but it should be empty.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input
1

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

大意：几个小镇要通过公路互通，已经建成了一些公路，问还须要建哪几条。并满足公路总长最小。

有多组输入，每组输入的格式是：n个城镇的坐标（x。y）  m条已经修好的路（公路一端城镇编号，还有一端编号）。

输出格式为：（待建公路一端城镇编号，还有一端编号）特殊判题，不用在意输出顺序。

注意：相邻两组測试例子的输出要有空行。最后一组后不加空行。不然就WR，居然还不是PE。zoj不够友好。

//zoj-2048-ac  special judge
#include
#include
#include
#define M 750*750/2
using namespace std;
int n,m;//共n个点，已有m条边
struct edge {
	int v1,v2;
	double dist;
};
struct vertice{
	int  x,y;
};
vertice arr_v[M];
priority_queue pq;
int tree[M];
bool operator<(const edge&a,const edge&b){
	if(a.dist>b.dist) return true;//保证pq.top()是当前最小的
	return false;
}
int f_find_root(int x){
	if(tree[x]==-1)
	  return x;
	else{
		int tmp=f_find_root(tree[x]);
		tree[x]=tmp;
		return tmp;
	}
}//f_find_root
void f_union(int a,int b){
	a=f_find_root(a);
	b=f_find_root(b);
	if(a!=b)  tree[b]=a;
}//f_union
void f_init(){
		for(int i=0;i>n;vertice tmpv;
	for(int i=0;i>arr_v[i].x>>arr_v[i].y;
	cin>>m;
	int v1,v2;
	for(int i=0;i>v1>>v2; v1--;v2--;
		f_union(v1,v2);
	}
}
void f_calc(){
	edge tmpe;double dx,dy;
	int v1,v2;
	for(int i=0;i>ncase;
    while(ncase--){
    f_init();
	f_in();
	f_calc();
	if(ncase) cout<