poj 2421 Constructing Roads 最小生成树(已经存在几条边) KRUSKAL

最新推荐文章于 2019-07-24 16:27:36 发布

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本文介绍了一个构造道路问题，目标是最小化连接所有村庄所需道路的总长度。通过输入村庄间距离及已建成道路信息，利用算法寻找最优解。

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13066		Accepted: 5221

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

Sample Output

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int fath[110];
int r[11000],u[11000],v[11000],w[11000];
int map[110][110];
int used[110][110];
int find(int x)
{
    return fath[x]==x?fath[x]:fath[x]=find(fath[x]);
}
bool cmp(int i,int j)
{
    return w[i]<w[j];
}
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        memset(used,0,sizeof(used));
        for(int i=1;i<=n;i++) fath[i]=i;
        for(int i=1;i<=n*(n-1)/2;i++) r[i]=i;
        for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) scanf("%d",&map[i][j]);
        int Q;scanf("%d",&Q);
        while(Q--)
        {
            int x,y;scanf("%d%d",&x,&y);
            used[x][y]=used[y][x]=1;
        }
        int tmp=1;//可用边数
        for(int i=1;i<=n;i++)
        {
            for(int j=i;j<=n;j++)
            {
                if(map[i][j]==0) continue;
                if(used[i][j])
                {
                    int x=find(i),y=find(j);
                    fath[x]=y;
                    continue;
                }
                u[tmp]=i,v[tmp]=j,w[tmp++]=map[i][j];
            }
        }
        sort(r+1,r+tmp,cmp);
        int cnt=0;
        for(int i=1;i<tmp;i++)
        {
            int e=r[i];
            int x=find(u[e]),y=find(v[e]);
            if(x!=y)
            {
                cnt+=w[e];
                fath[x]=y;
            }
        }
        printf("%d/n",cnt);
    }
    return 0;
}