Luogu P5098 Cave Cows 3

题目大意

  有(n)个坐标((x_i, y_i))，求任意两个坐标之间的最大曼哈顿距离。
  (1 leq n leq 50000)，(-1 imes 10^6 leq x_1,x_2,y_1,y_2 leq 1 imes 10^6)。
 

题解

  我们把(|x_1 - x_2| + |y_1 - y_2|)分成四种情况：

[egin{aligned} &|x_1 - x_2| + |y_1 - y_2| = x_1 - x_2 + y_1 - y_2 = (x_1 + y_1) - (x_2 + y_2) &(x_1 geq x_2, y_1 geq y_2) \ &|x_1 - x_2| + |y_1 - y_2| = x_1 - x_2 + y_2 - y_1 = (x_1 - y_1) - (x_2 - y_2) &(x_1 geq x_2, y_1 < y_2) \ &|x_1 - x_2| + |y_1 - y_2| = x_2 - x_1 + y_1 - y_2 = -((x_1 - y_1) - (x_2 - y_2)) &(x_1 < x_2, y_1 geq y_2) \ &|x_1 - x_2| + |y_1 - y_2| = x_2 - x_1 + y_2 - y_1 = -((x_1 + y_2) - (x_2 + y_2)) &(x_1 < x_2, y_1 < y_2) end{aligned}]

  显然分别求(x_i - y_i)的最大值和最小值即可。

#include <iostream>
#include <cstdio>

using namespace std;

int n;
int x, y;
int maxp = -(1 << 30), minp = 1 << 30, maxd = -(1 << 30), mind = 1 << 30;

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d%d", &x, &y);
        minp = min(minp, x + y);
        maxp = max(maxp, x + y);
        mind = min(mind, x - y);
        maxd = max(maxd, x - y);
    }
    printf("%d", max(maxp - minp, maxd - mind));
    return 0;
}

  当然我们把曼哈顿距离转化成切比雪夫距离。就是把每个

[(x_i, y_i) ]

  转化成

[(x_i + y_i, x_i - y_i) ]

  那么曼哈顿距离

[|x_1 - x_2| + |y_1 - y_2| ]

  就可以转化成对应的切比雪夫距离

[max { |(x_1 + y_1) - (x_2 + y_2)|, |(x_1 - y_1) - (x_2 - y_2)| } ]

  可以发现代码都是一样的。