hdu1102Constructing Roads

Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28201    Accepted Submission(s): 10744 http://acm.hdu.edu.cn/showproblem.php?pid=1102   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：

1到N个村庄，已知一些村庄之间已经有一些道路。要求修建一些道路，使所有村庄都连通起来，使所有道路长度最小。
 

思路：

最小生成树问题(Prim算法)。  把之前就有道路的村庄看成一个点，即mapp[i][j]=mapp[j][i]=0。

代码：

#include<iostream>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;

int vis[105];
int mapp[105][105];
int l[105][105];
int dis[105];
int n;

int prim(int s)
{
    long long sum=0;
    vis[s]=0;
    for(int i=1;i<=n;i++)
        dis[i]=i==s?0:mapp[i][s];
    int minn;
    int k;
    for(int i=1;i<=n;i++)
    {
        minn=inf;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&minn>dis[j])
            {
                minn=dis[j];
                k=j;
            }
        }
        vis[k]=1;
        sum+=minn;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dis[j]>mapp[k][j])
                dis[j]=mapp[k][j];
        }
    }
    return sum;
}

int main()
{
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                cin>>mapp[i][j];
        }
        int q;
        cin>>q;
        int a,b;
        memset(vis,0,sizeof(vis));
        while(q--)
        {
            cin>>a>>b;;
            mapp[a][b]=mapp[b][a]=0;
        }
        long long sum=prim(1);
        cout<<sum<<endl;
    }
    return 0;
}