HDOJ  4312   Meeting point-2

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

题解报告;

平面上两点间的 Chebyshev距离为 max(|x1-x2|, |y1-y2|)

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求max(|x1-x2|, |y1-y2|)

怎么样才能将这一道题转化为上一道求解，

2*max(|x1-x2|, |y1-y2|)=|x1-x2-(y1-y2)|+|x1-x2+(y1-y2)|=|(x1-y1)-(x2-y2)|+|(x1-y1)+(x2+y2)|

上一题解的是 

来那个这相比较

可以将x1=x1-y1,x2=x2-y2;y1=x1-y1,y2=x2+y2;对待

所以将上一题的答案改一下就好

#include <iostream>
 #include <algorithm>
 #include <cstdio>
 using namespace std;
 typedef long long ll;
 const int MAXN = 100100;
 struct Point{
     int x, y;
     Point(){}
     Point(int _x, int _y):x(_x),y(_y){}
 }ary[MAXN];
 int x[MAXN], y[MAXN];
 ll sx[MAXN], sy[MAXN];
 int main()
 {
     int T, n;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         for(int i = 0; i < n; ++i)
         {
             scanf("%d%d",&x[i],&y[i]);
             ary[i] = Point(x[i]-y[i],x[i]+y[i]);        //改动
             x[i]=ary[i].x,y[i]=ary[i].y;        //改动
         }
         sort(x, x + n);
         sort(y, y + n);
         sx[0] = x[0];
         sy[0] = y[0];
         for(int i = 1; i < n; ++i)
         {
             sx[i] = sx[i - 1] + x[i];
             sy[i] = sy[i - 1] + y[i];
         }
         ll ans = 1;
         ans <<= 60;
         for(int i = 0; i < n; ++i)
         {
             int rankx = lower_bound(x, x + n, ary[i].x) - x;
             int ranky = lower_bound(y, y + n, ary[i].y) - y;
                           +sx[n - 1] - sx[rankx] - ll(ary[i].x) * (n - rankx - 1);
                            +sy[n - 1] - sy[ranky] - ll(ary[i].y) * (n - ranky - 1);
             ans = min(ans, t);
         }
         cout<<ans/2<<endl;
     }
     return 0;
 }