poj

Constructing Roads

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 25409 Accepted: 11102

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

30 990 692990 0 179692 179 011 2

Sample Output

179

#include<iostream>#include<cstdio>int main(){    int n,q,a[105][105],sum=0;    int flag[105]={0};    scanf("%d",&n);    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            scanf("%d",&a[i][j]);        }    }    scanf("%d",&q);    int b,c;    for(int i=1;i<=q;i++){        scanf("%d%d",&b,&c);        a[b][c]=0;        a[c][b]=0;    }    flag[1]=1;//第一个结点被覆盖掉    for(int k=1;k<n;k++){//循环n-1次找到n个结点就可        int min1=-1,min_i;        for(int i=2;i<=n;i++){//选取下一个最小权值的结点            if(flag[i]==0&&(min1==-1||a[1][i]<min1)){                min1=a[1][i];                min_i=i;//记录被选的结点号            }        }        flag[min_i]=1;        for(int i=2;i<=n;i++){//更新未覆盖结点距离            if(flag[i]==0&&a[1][i]>a[min_i][i]){                a[1][i]=a[min_i][i];            }        }        sum+=a[1][min_i];//加上权值    }    printf("%d\n",sum);}

这道题多了一个已建立的两个村庄之间的路（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.）已建立的村庄就不用我们再建了，所以就赋值该两个村庄的距离为0，不算到mininum里。