POJ 2421 Constructing Roads

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24967		Accepted: 10857

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

n个村庄，且两两村庄之间距离已经给出，接下来q条，表明村庄之间已经建立链接，清零就可以了；

//最小生成树
#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath> 
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f 
#define mem(a) (memset(a,0,sizeof(a))) 
typedef long long ll;
int g[150][150];
int vis[150],n,m;
int dis[150],x,y;
int prime()
{
    for(int i=1;i<=n;i++)
    {
        dis[i]=g[1][i];
        vis[i]=0;
    }
    vis[0]=1;
    int v,minn,sum=0;
    for(int i=1;i<=n;i++)
    {
        minn=INF;v=0;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && minn>dis[j])
            {
                v=j;
                minn=dis[j];
            }
        }
        vis[v]=1;
        if(minn!=INF) sum+=minn;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]) dis[j]=min(dis[j],g[v][j]);
        }
    }
    return sum;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&g[i][j]);
        }
    }
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d%d",&x,&y);
        g[x][y]=g[y][x]=0;
    }
    printf("%d
",prime());
    return 0;
}