poj2421Constructing Roads-最小生成树prim

                                                                                                                Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16680		Accepted: 6806

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

 

//汗，一连做几个题目都是最小生成树的，总的思想不变，格式稍稍变化，而对于这题的话，就是有些路已经建好了的，不需要你考虑在内的，那么就把已经建好了的初始化一下~还有就是无向图。所以map[a][b]=map[b][a]=0;

水水的代码：

#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define inf 9999
int vis[1001];
int map[101][101];
int dis[101];
int n;
int Q;
int a,b;
int Prim(){
	memset(vis,0,sizeof(vis));
	int sum=0; int k;
	for(int i=1;i<=n;i++)
		dis[i]=map[1][i];
	vis[1]=1;   
	for(int i=1;i>n){
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				cin>>map[i][j];
		cin>>Q;
		for(int k=1;k<=Q;k++){
			cin>>a>>b;
			map[a][b]=map[b][a]=0;
		}
		cout<