​​http://www.elijahqi.win/archives/372​​​
首先提醒一下，洛谷上a的poj AC不了因为Poj多组测试数据

这题是2016年11月12日写的，当时没写过，就放着了，那时应该刚考完noip 其实自己连最小生成树都不会写，抱着210的分数竟然还拿了天津的省一，有点可笑

注意一下，data也就是结构体要开大，我当时错就在这，今天也花了不少时间才查出来。。

stl大概比手写快排慢，这题不疏密其实prime会更好，但是我暂时只会kruskal，就只写了这个得

Description

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.
Input

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.
Output

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.
Sample Input

4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0
Sample Output

28
Source

USACO 102

#include<cstdio>
struct node{
    int x,y,z;
};
node data[11000];
int n,a,aa[110][110],f[110];
void qsort(int l,int r){
    int i=l,j=r,mid=data[(l+r)>>1].z;
    while (i<=j){
        while (data[i].z<mid)++i;
        while (data[j].z>mid)--j;
        if (i<=j){
            node t=data[i];data[i]=data[j];data[j]=t;++i;--j;
        }
    } 
    if (l<j) qsort(l,j);
    if (i<r) qsort(i,r);
}
inline int find(int x){
    return f[x]==x?x:f[x]=find(f[x]);
} 
int main(){
    //freopen("p1546.in","r",stdin);
    //freopen("p1546.out","w",stdout);
    while (~scanf("%d",&n)){
        int num=0;
        for (int i=1;i<=n;++i){
            for (int j=1;j<=n;++j) scanf("%d",&aa[i][j]);
        }
        for (int i=2;i<=n;++i){
            for (int j=1;j<=i-1;++j){
                data[++num].x=i;
                data[num].y=j;
                data[num].z=aa[i][j];
            }
        }
        //for (int i=1;i<=num;++i) printf("%d,data[i].z);
        qsort(1,num);
        //for (int i=1;i<=num;++i) printf("%d,data[i].z);
        for (int i=1;i<=n;++i) f[i]=i;
        int ans=0;
        int nn=0;
        for (int i=1;i<=num;++i){
            int x=data[i].x,y=data[i].y;
            int xx=find(x),yy=find(y);
            if (xx!=yy){
                ++nn;f[xx]=f[yy];ans+=data[i].z;
            }else continue;
            if (nn==(n-1)) break;
        }    
        printf("%d\n",ans);
    }
    return 0;
}