hdu-1102 Constructing Roads(最小生成树)

题目链接：

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 65536/32768 K (Java/Others)

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：

 

给出一个矩阵代表两个点之间的距离,现在要求把所有的点连起来,其中已经有一些点连起来了,问最少还要建多长的路才能把所有点连起来;

 

思路：

 

最小生成树的模板题;

 

AC代码：

/*1102    15MS    1684K    1365 B    G++    2014300227*/
#include <bits/stdc++.h>
using namespace std;
const int N=103;

typedef long long ll;
const ll mod=1e9+7;
const double PI=acos(-1.0);
int n,a[103][103],p[103],q,u,v;
int findset(int x)
{
    if(x==p[x])return x;
    return p[x]=findset(p[x]);
}
void same(int x,int y)
{
    int fx=findset(x),fy=findset(y);
    if(fx!=fy)p[fx]=p[fy];
}
struct Edge
{
    int l,r,len;
};
Edge edge[5010];
int cmp(Edge x,Edge y)
{
    return x.len<y.len;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            p[i]=i;
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&a[i][j]);
                if(j>i)
                {
                    edge[cnt].l=i;
                    edge[cnt].r=j;
                    edge[cnt++].len=a[i][j];
                }
            }
        }
        sort(edge,edge+cnt,cmp);
        scanf("%d",&q);
        for(int i=1;i<=q;i++)
        {
            scanf("%d%d",&u,&v);
            same(u,v);
        }
        int ans=0;
        for(int i=0;i<cnt;i++)
        {
            if(findset(edge[i].l)!=findset(edge[i].r))
            {
                ans+=edge[i].len;
                same(edge[i].l,edge[i].r);
            }
        }
        printf("%d
",ans);

    }

    return 0;
}