POJ 2421 Constructing Roads

题目链接

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28254		Accepted: 12483

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：有N个村庄，编号1~N。现需要在这N个村庄之间修路，使得任何两个村庄之间可以连通。已知某些村庄之间已经有道路连接，试修建一些路使得所有村庄都是连通的、且修路总长最短

AC代码：

#include <iostream>
#include <algorithm>
using namespace std;
const int N=110;
struct node{
    int u,v,d;
}p[N*N];
int pre[N];
int n,cnt;
void init(){
    for(int i=1;i<=n;i++){
        pre[i]=i;
    }
}
int Find(int x){
    if(x!=pre[x]) pre[x]=Find(pre[x]);
    return pre[x];
}
void join(int x,int y){
    int tx=Find(x);
    int ty=Find(y);
    if(tx!=ty){
        pre[ty]=tx;
    }
}
bool cmp(node a,node b){
    return a.d<b.d;
}
int Kru(){
    int ans=0;
    sort(p,p+cnt,cmp);
    for(int i=0;i<cnt;i++){
        if(Find(p[i].u)!=Find(p[i].v)){
            join(p[i].u,p[i].v);
            ans+=p[i].d;
        }
    }
    return ans;
}
int main(){
    int q,a,b;
    while(~scanf("%d",&n)){
        cnt=0;
        init();
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&a);
                if(i==j) continue;
                p[cnt].u=i;
                p[cnt].v=j;
                p[cnt++].d=a;
            }
        }
        scanf("%d",&q);
        while(q--){
            scanf("%d%d",&a,&b);
            p[cnt].u=a;
            p[cnt].v=b;
            p[cnt++].d=0;
        }
        int ans=Kru();
        printf("%d
",ans);
    }
    return 0;
}