HDU-1102: Constructing Roads

最小生成树

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：输入一个数字n，再输入n行n列数字，i行j列代表第i个村庄到第j个村子的距离。再输入一个整数q，下面有q行，每行两个数字a和b，表示村庄a和b之间的路已经修好了，问连通所有的村子的还需要的最小长度。

分析：最小生成树模板题。首先输入m[i][j]的值，然后q组已经修好的路，令m[a][b]为0即可。

• 完整代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
int m[111][111],father[11111],k,sum;
struct node
{
    int st,en,len;
}a[11111];
int cmp(node a,node b)
{
    return a.len<b.len;
}
int findx(int x)
{
    if(x!=father[x])
        father[x]=findx(father[x]);
    return father[x];
}
void kruskal()
{
    int i,x,y;
    sum=0;
    for(i=0;i<k;i++)
        father[i]=i;
    for(i=0;i<k;i++)
    {
        x=findx(a[i].st);
        y=findx(a[i].en);
        if(x!=y)
        {
            sum+=a[i].len;
            father[x]=y;
        }
    }
    printf("%d\n",sum);
}
int main()
{
    int n,i,j,x,y,q;
    while(scanf("%d",&n)!=EOF)
    {
        k=0;
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                scanf("%d",&m[i][j]);
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d%d",&x,&y);
            m[x][y]=0;
            m[x][y]=0;
        }
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
        {
            a[k].st=i;
            a[k].en=j;
            a[k].len=m[i][j];
            k++;
        }
        sort(a,a+k,cmp);
        kruskal();
    }
    return 0;
}