Constructing Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13995 Accepted Submission(s): 5324

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

Source

这道题和之前做的那道题类似，直接用prim再次水过，思想也和那道题类似，就是把两个已经连通的点之间的距离设为0即可了，然后再直接用prim水过；

以下是代码：

#include 
      
       #include 
       
        using namespace std;const int maxn=101;const int Max=0x3f3f3f3f;int map[maxn][maxn],low[maxn],visit[maxn];int n;int prim(){    int i,j,pos,min,mst=0;    memset(visit,0,sizeof(visit));    pos=1;    visit[1]=1;    for(i=1;i<=n;i++)        low[i]=map[pos][i];    for(i=1;i
        
         low[j])            {                min=low[j];                pos=j;            }        }        mst+=min;        if(mst>=Max) break;//说明这个图不连通        visit[pos]=j;        for(j=1;j<=n;j++)        {            if(!visit[j] && low[j]>map[pos][j])//更新low数组                low[j]=map[pos][j];        }    }    return mst;}int main(){    int q,i,j;    int a,b;    while(scanf("%d",&n)!=EOF)    {        memset(map,Max,sizeof(map));        for(i=1;i<=n;i++)          for(j=1;j<=n;j++)            scanf("%d",&map[i][j]);        scanf("%d",&q);       while(q--)       {           scanf("%d%d",&a,&b);           map[a][b]=map[b][a]=0;       }       printf("%d\n",prim());    }    return 0;}

你可能感兴趣的文章