HDU 1102Constructing Roads（最小生成树）

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22419    Accepted Submission(s): 8619

Problem Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

 

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

 

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

 

Sample Input

  
  

   
   3
0 990 692
990 0 179
692 179 0
1
1 2
  
  

 

Sample Output

  
  

   
   179
  
  

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m;
int per[102]; 
struct dis{
	int x,y,z;
}d[5055];
int find(int r){
	if(per[r]==r) return r;
	return per[r]=find(per[r]);
} 
void join(int r1,int r2){
	int f1=find(r1);
	int f2=find(r2);
	if(f1!=f2) per[f1]=f2;
}
bool cmp(dis d1,dis d2){
	return d1.z<d2.z;
} 
int main(){
	
	while(scanf("%d",&n)!=EOF&&n){
	int dsum=0;//需要的修建的距离 
  	int ans=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			int k;scanf("%d",&k);
			if(j>i){
				d[ans].x=i;d[ans].y=j;d[ans].z=k;ans++;
			}
		}
	} 
	for(int i=1;i<=n;i++){
		per[i]=i;//第一步：将每个顶点形成独立的树 
	} 
    scanf("%d",&m);
	for(int i=0;i<m;i++){
		int x1,y1;
		scanf("%d%d",&x1,&y1);
		join(x1,y1);//将已经连接的合并成一棵树 
	}
	sort(d,d+ans,cmp);//第二步：将边从小到大排序
	for(int i=0;i<ans;i++){//第三步：判断边的两个端点是否属于同一个树，如果不是，就把它们合并，并更新最少的距离 
		if(find(d[i].x)!=find(d[i].y)){
			join(d[i].x,d[i].y);
			dsum+=d[i].z; 
		} 
	}
	printf("%d\n",dsum);
}
	
	return 0;

}