POJ2421 Constructing Roads

 

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23071		Accepted: 9894

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

Source

PKU Monthly,kicc

 

给出一个邻接矩阵，再给出Q条已经连好的边，求最小生成树中最大边的距离。

裸最小生成树。

 

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=3210;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct edge{
    int x,y;
    int d;
}e[mxn*100];
int cmp(const edge a,const edge b){
    return a.d<b.d;
}
int m,n;
//
int fa[mxn];
int find(int x){
    if(fa[x]==x)return x;
    return fa[x]=find(fa[x]);
}
int cnt=0;
void kruskal(){

    int num=1;
    int ans=0;
    for(int i=1;i<=cnt;++i){
        int f1=find(e[i].x);
        int f2=find(e[i].y);
        if(f1!=f2){
            fa[f1]=f2;
            ans+=e[i].d;
//            printf("%d %d %d\n",e[i].x,e[i].y,ans);
            num++;
        }
        if(num==cnt-m-1)break;
    }
    printf("%d\n",ans);
    return;
}
int T;
int main(){
    int i,j;
    n=read();
    cnt=0;
    for(i=1;i<=n;i++)
     for(j=1;j<=n;j++){
         m=read();
         if(i==j)continue;
         e[++cnt].x=i;e[cnt].y=j;
         e[cnt].d=m;
//         printf("  %d %d %d\n",i,j,m);
     }
    sort(e+1,e+cnt+1,cmp);
    int u,v;
    m=read();
    for(int i=1;i<=n;++i)fa[i]=i;
    for(i=1;i<=m;++i){
        u=read();v=read();
        u=find(u);v=find(v);
        if(u!=v)fa[u]=v;
    }
    kruskal();
    return 0;
}