hdu 1102 Constructing 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): 19671    Accepted Submission(s): 7508

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
   
   
   
   

       题意：已知有N个村庄，要求的是需要修最短的路使从任意一个村庄都能到达其他任意的村庄。其中有Q条路是已经存在的，给出了村庄到村庄之间的距离矩阵。 最小生成树问题，prim算法。

代码：

#include<iostream>
using namespace std;
int dis[110][110];  //村庄之间的距离矩阵 
int v1,v2;
int mincost[110];  //从集合X出发的边到每个顶点的最小权值 
int ans;
int N;
void prim()
{
    int i,j;
    for(i=1;i<=N;i++)
    {
        if(i!=v1)
        {
            mincost[i]=dis[v1][i];
        }
    }
    mincost[v1]=-1;
    int flag=v1;
    for(i=1;i<N;i++)
    {
        int Min=10000;
        for(j=1;j<=N;j++)
        {
            if(Min>mincost[j] && mincost[j]>=0)   //找到最短路 
            {
                Min=mincost[j];
                flag=j;
            }
        }
        ans+=mincost[flag];   //更新需要修的路的长度 
        mincost[flag]=-1;     //用过的路标记为负数 
        for(j=1;j<=N;j++)
        {
            if(dis[flag][j]<mincost[j])
            {
                mincost[j]=dis[flag][j];
            }
        }
    }
}
int main()
{
    while(cin>>N)
    {
        int i,j;
        for(i=1;i<=N;i++)   //距离矩阵 
        {
            for(j=1;j<=N;j++)
            {
                cin>>dis[i][j];
            }
        }
        int Q;
        cin>>Q;
        for(i=1;i<=Q;i++)  //对于已有的路，标记两村庄之间的距离为0 
        {
            cin>>v1>>v2;
            dis[v1][v2]=0;    
            dis[v2][v1]=0;
        }
        ans=0;
        prim();
        cout<<ans<<endl;
    }
    return 0;
}