poj2421解题报告

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12166		Accepted: 4831

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

题意：n个乡村间要修路，已经修好了一部分的路，给出每两个村庄间的距离，求剩下的路最少要修多长

思路：最小生成树的小变种，如果将已经存在的路距离修改成0，这样就会保证用prim生成树时这条路一定会入选并且

长度不会计算在最小生成树内。。。。

#include
using namespace std;
int main()
{
    int n,i,j,distance[101][101],lock[101],point,a,b,s;
    while(cin>>n&&n)
    {
        memset(lock,0,sizeof(lock));
        int mintree=0,finded=0,min;
        for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
				cin>>distance[i][j];
			cin>>s;
			if(s)
			{
				while(s--)
				{
					cin>>a>>b;
					distance[a][b]=distance[b][a]=0;
				}
			}
			finded=1;
			lock[1]=1;
            while(finded!=n)
            {
                min=100000000;
                for(i=1;i<=n;i++)
                    for(j=1;j<=n;j++)
                        if(lock[i]==1&&lock[j]==0)
                            if(min>distance[i][j])
                            {
                                min=distance[i][j];    
                                point=j;
                            }
                            
							mintree+=min;    lock[point]=1;            
							finded++;            
            }
			cout<