[USACO2004OPEN]Cave Cows 3

嘟嘟嘟

看完题后突然想起jf巨佬的话：“看到曼哈顿距离就想转切比雪夫距离。”

于是我就转换了一下。

然后问题变成了求

\[max_{i, j \in n} \{ max \{ |x_i - x_j|, |y_i - y_j| \} \}
\]

令差最大，只要分别找出\(x, y\)的最小值和最大值即可。

\(O(n)\)扫一遍啦。

#include<cstdio>

#include<iostream>

#include<cmath>

#include<algorithm>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<vector>

#include<stack>

#include<queue>

using namespace std;

#define enter puts("")

#define space putchar(' ')

#define Mem(a, x) memset(a, x, sizeof(a))

#define In inline

typedef long long ll;

typedef double db;

const int INF = 0x3f3f3f3f;

const db eps = 1e-8;

const int maxn = 5e4 + 5;

inline ll read()

{

	ll ans = 0;

	char ch = getchar(), last = ' ';

	while(!isdigit(ch)) {last = ch; ch = getchar();}

	while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}

	if(last == '-') ans = -ans;

	return ans;

}

inline void write(ll x)

{

	if(x < 0) x = -x, putchar('-');

	if(x >= 10) write(x / 10);

	putchar(x % 10 + '0');

}

int n;

struct Node

{

	int x, y;

}t[maxn];

int main()

{

	n = read();

	for(int i = 1; i <= n; ++i)

	{

		int x = read(), y = read();

		t[i] = (Node){x + y, x - y};

	}

	int Max = 1, May = 1, Mix = 1, Miy = 1;

	for(int i = 2; i <= n; ++i)

	{

		if(t[i].x < t[Mix].x) Mix = i;

		if(t[i].x > t[Max].x) Max = i;

		if(t[i].y < t[Miy].y) Miy = i;

		if(t[i].y > t[May].y) May = i;

	}

	write(max(t[Max].x - t[Mix].x, t[May].y - t[Miy].y)), enter;

	return 0;

}

巴特西

[USACO2004OPEN]Cave Cows 3

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