题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4312

题意：给定平面坐标上n（n<=100000）个点，然后在其中选一个，使得所有点到当前点的Chebyshev距离和最小。

分析：

切比雪夫距离：设a(x1,y1),b(x2,y2);DIS = max(|x1-x2|,|y1-y2|) = (|x1-x2+y1-y2|+|x1-x2-y1+y2|)/2;

我们将点aa的坐标看成（x1+y1,x1-y1）,bb的坐标看成（x2+y2,x2-y2）,从几何意义上讲相当于点在原

坐标系上逆时针旋转45度。

然后求新的的最小的曼哈顿距离和的一半即可。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long LL;
const int maxn = 1e5+10;

struct point{
    int x,y;
    LL sum;
}p[maxn];

bool cmp1(point A,point B)
{
    if(A.x<B.x) return true;
    if(A.x==B.x&&A.y<B.y) return true;
    return false;
}

bool cmp2(point A,point B)
{
    if(A.y<B.y) return true;
    if(A.y==B.y&&A.y<B.y) return true;
    return false;
}


int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        LL sum=0;
        int x,y;
        for(int i=1;i<=n;i++){
            scanf("%d%d",&x,&y);
            p[i].x=x-y;
            p[i].y=x+y;
            p[i].sum=0;
        }
        sort(p+1, p+1+n, cmp1);
        for (LL i = 1; i <= n; ++i) {
            p[i].sum += (i-1) * p[i].x - sum;
            sum += p[i].x;
        }
        sum = 0;
        for (LL i = n; i >= 1; --i) {
            p[i].sum += sum - (n-i) * p[i].x;
            sum += p[i].x;
        }
        sum = 0;
        sort(p+1, p+1+n, cmp2);
        for (LL i = 1; i <= n; ++i) {
            p[i].sum += (i-1) * p[i].y -sum;
            sum += p[i].y;
        }
        sum = 0;
        LL ans = 1LL<<62;
        for (LL i = n; i >= 1; --i) {
            p[i].sum += sum - (n-i) * p[i].y;
            ans = min(ans, p[i].sum);
            sum += p[i].y;
        }
        printf("%I64d\n",ans/2);
    }
    return 0;
}

HDU4312 Meeting point-2 (切比雪夫距离&&曼哈顿距离)