Built?

There are N towns on a plane. The i-th town is located at the coordinates (xi,yi). There may be more than one town at the same coordinates.
You can build a road between two towns at coordinates (a,b) and (c,d) for a cost of min(|a−c|,|b−d|) yen (the currency of Japan). It is not possible to build other types of roads.
Your objective is to build roads so that it will be possible to travel between every pair of towns by traversing roads. At least how much money is necessary to achieve this?

Constraints
2≤N≤105
0≤xi,yi≤109
All input values are integers.

 

输入

Input is given from Standard Input in the following format:

N
x1 y1
x2 y2
:
xN yN

 

输出

Print the minimum necessary amount of money in order to build roads so that it will be possible to travel between every pair of towns by traversing roads.

 

样例输入

3
1 5
3 9
7 8

样例输出

3 

提示

Build a road between Towns 1 and 2, and another between Towns 2 and 3. The total cost is 2+1=3 yen.

这个题目大致意思就是给了n个点的坐标，最终目的是这N个点两两可以走到，两个点的权值是min(fabs(x1-x2),fabs(y1-y2));

这其实就是就是一个最小生成树的变形；

因为一个点只会和离他最近的两个点进行连接，所以要分别按照x和Y进行排序，从而确定权值，然后再用并查集原理解决就好了；

下面是AC代码

#include <bits/stdc++.h>
using namespace std;
//#define mod 1000000007
int fa[110000];
typedef struct
{
    int x,y,hao;//x坐标，y坐标，和对应序号
}qq;
typedef struct
{
    int p1,p2,p3;//分别是两个点之间的距离和，两个点所对应标号
}pp;
bool cmp1(qq r,qq t)
{
    return r.x<t.x;//按照X排序
}
bool cmp2(qq r,qq t)
{
    return r.y<t.y;//按照y排序
}
bool cmp3(pp r,pp t)
{
    return r.p1<t.p1;//最后的路径的权值排序
}
int getfa(int a)//找父节点的函数
{
    if(fa[a]!=a)
    {
        return fa[a]=getfa(fa[a]);
    }
    else
    {
        return a;
    }
}
qq a[110000];
pp b[440000];
int main()
{
    int n;
     scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        fa[i]=i;
    }
 for(int i=0;i<n;i++)
    {
        scanf("%d%d",&a[i].x,&a[i].y);
        a[i].hao=i;
    }
    sort(a,a+n,cmp1);
    int k=0;
    for(int i=1;i<n;i++)
    {
       b[k].p1=a[i].x-a[i-1].x;
       b[k].p2=a[i].hao;
       b[k].p3=a[i-1].hao;
       k++;
    }
    sort(a,a+n,cmp2);
    for(int i=1;i<n;i++)
    {
        b[k].p1=a[i].y-a[i-1].y;
        b[k].p2=a[i].hao;
        b[k].p3=a[i-1].hao;
        k++;
    }
    sort(b,b+2*n-2,cmp3);
    long long int s=0;//要用long long int 
    for(int i=0;i<2*n-2;i++)
    {
        if(getfa(b[i].p2)!=getfa(b[i].p3))
        {
            s+=b[i].p1;
            fa[getfa(b[i].p3)]=getfa(b[i].p2);
        }
    }
    printf("%lld\n",s);
    return 0;
}