POJ2421Constructing Roads

　　　　　　　　　　　　　　　Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23343		Accepted: 10015

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

题目描述：

有N个村庄,从1到N编号,建立一些道路,这样任意两个村庄能连接到对方。我们说两个村庄A和B是连接,当且仅当有A和B之间的道路,或存在C这样的一个村庄之间有一条路A和C,C和B连接。

我们知道,已经有一些道路之间的一些村庄,你的工作是构建道路,所有的村庄都连接和道路建设是最小的长度。

输入描述：

第一行是一个整数N(3 < = N < = 100),这是村庄的数量。然后N行,其中包含N个整数,第i和j N个整数的距离(距离应该是整数在[1000])j村之间我和村庄。

还有一个整数(0 < = Q < = N *(N + 1)/ 2)。然后再问行,每一行包含两个整数a和b(1 < = < b < = N),这意味着路村和村b之间已经建立了。

输出描述：

你应该输出一行包含一个整数,它是所有道路的长度建立连接,这样所有的村庄,这个值是最低。

 

/*
很简单的最小生成树 
*/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{int x,y,t;}e[20010];
inline void read(int &x)
{
    int f=1;x=0;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    x=x*f;return ;
}
bool cmp(const node &a,const node &b){return a.t<b.t;}
int n,m,x,y,ans=0,fa[110],sum=0;
int find(int x)
{
    if(fa[x]==x)return x;
    return fa[x]=find(fa[x]);
}
void add(int x,int y)
{
    int xx=find(x),yy=find(y);
    fa[xx]=yy;
}
int main()
{
    read(n);
    for(int i=1;i<=n;i++)fa[i]=i;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {    
            sum++;
            read(e[sum].t);
            e[sum].x=i;e[sum].y=j;
        }
    read(m);
    for(int i=1;i<=m;i++)
    {
        read(x),read(y);
        e[++sum].x=x;
        e[sum].y=y;
        e[sum].t=0;
    }
    sort(e+1,e+sum+1,cmp);
    m=0;
    for(int i=1;i<=sum;i++)
    {
        int xx=find(e[i].x),yy=find(e[i].y);
        if(xx!=yy)
        {
            ans+=e[i].t;
            add(xx,yy);
            m++;
        }
        if(m==n-1)break;
    }
    printf("%d
",ans);
    return 0;
}