POJ-2421Constructing Roads,又是最小生成树，和第八届河南省赛的引水工程惊人的相似，并查集与最小生成树的灵活与能用，水过~~~

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K

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

找的最小生成树专题里的一个题，看到这题却发现和河南省第八届省赛的“引水工程”惊人的相似，反正那道题也没做出来，本打算先A了那道题再做这道题，可是做了那题之后又发现这道题还更简单，于是调调代码直接A了，其实我宁愿相信是后台测试数据水的；

来说说思路吧:我们发现输入是以矩阵的模式，如果全部存在结构体中无疑是浪费内存，我们发现这个矩阵中很多相同的数，即坐标相对应的点（如a[i][j]=a[j][i]）的值是一样的,所以只需要把矩阵中的一半的数存在结构体中就可以了，然后就是最简单的最小生成树了，至于有Q条已经联通好的路我们只需在输入得时候利用并查集将输入的数据的根节点合并即可；可能这里表达有点不清楚，来看代码就马上明白了；

AC代码：说白了，这题就是并查集与最小生成树的灵活运用；

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=5100;//数据范围是100，平方是10000，但Q<=10000/2,开5050就可以过了；
int n,m,k,f[100];
struct node
{
    int u,v,w;
} a[N];
int cmp(node a,node b)
{
    return a.w<b.w;
}
int find(int x)
{
    return f[x]==-1?x:x=find(f[x]);
}
int ks(int n)
{
    int ans=0,cot=0;
    sort(a+1,a+1+k,cmp);
    for(int i=1; i<=k; i++)
    {
        int u=find(a[i].u);
        int v=find(a[i].v);
        if(u!=v)
        {
            ans+=a[i].w;
            f[u]=v;
            cot++;
        }
        if(cot==n-1)
            break;
    }
    return ans;
}
int main()
{
    int x;
    scanf("%d",&n);
    k=0;
    memset(a,0,sizeof(a));
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
        {
            scanf("%d",&x);
            if(j>i)//把一半的数据存在结构体中；
                a[++k].u=i,a[k].v=j,a[k].w=x;
        }
    memset(f,-1,sizeof(f));
    scanf("%d",&m);
    int  x1,y1;
    while(m--)
    {
        scanf("%d%d",&x1,&y1);
        int xx=find(x1);
        int yy=find(y1);
        if(xx!=yy)//根节点合并->并查集；
        {
            f[xx]=yy;
            n--;
        }
    }
    printf("%d\n",ks(n));
    return 0;
}