Constructing Roads

绝地灬酷狼 2022-06-11 03:38 115阅读 0赞

# Constructing Roads #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 23858    Accepted Submission(s): 9168**

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：就是求最小生成树,只需要注意有些是已经连通了的，这些将其赋值为0即可，其他就是套模板了
      
       
      
       
       
        #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    using namespace std;
    int map1[101][101];
    bool vis[101];
    int dis[101],n,ans;
    const int inf=0x3f3f3f3f;
    void prime()
    {
        int k;
        for(int i=2;i<=n;i++)
            dis[i]=map1[1][i];
        dis[1]=0;
        vis[1]=true;
        for(int i=1;i<n;i++)
        {
            int minn=inf;
            for(int j=1;j<=n;j++)
            {
                if(!vis[j]&&minn>dis[j])
                {
                    minn=dis[j];
                    k=j;
                }
            }
            ans+=minn;
            vis[k]=true;
            for(int j=1;j<=n;j++)
            {
                if(!vis[j]&&dis[j]>map1[k][j])
                    dis[j]=map1[k][j];
            }
        }
    }
    int main()
    {
        int m,x,y;
        while(~scanf("%d",&n))
        {
            ans=0;
            memset(vis,false,sizeof(vis));
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    scanf("%d",&map1[i][j]);
                }
            }
            scanf("%d",&m);
            for(int i=1;i<=m;i++)
            {
                scanf("%d %d",&x,&y);
                map1[x][y]=map1[y][x]=0;
            }
            prime();
            printf("%d\n",ans);
        }
        return 0;
    }