AtCoder Beginner Contest 215 F - Dist Max 2

平面上有一系列的点(xi,yi)(x_i,y_i)(xi,yi)，定义两点(xi,yi),(xj,yj)(x_i,y_i),(x_j,y_j)(xi,yi),(xj,yj)的距离为min(∣xi−xj∣,∣yi−yj∣)min(|x_i-x_j|,|y_i-y_j|)min(∣xi−xj∣,∣yi−yj∣)，求出最远的两个点的距离

想过用二分答案，但是没有想到怎么check，赛后看到题解发现其实很简单。

首先还是二分答案，要找到一个点对使两个元素的差值都要大于等于这个答案。所以，首先先对x坐标进行排序，那么点对的范围就可以容易找到，比如对于x1x_1x1有着xk−1<x1+ansx_{k-1}<x_1+ansxk−1<x1+ans并且xk>=x1+ansx_k>=x_1+ansxk>=x1+ans，那么k到n的点对就满足了x方向的条件。这个时候，我们维护一个后缀的最大值和最小值，O(1)O(1)O(1)查找k~n的y方向的最大值和最小值，和y1y_1y1进行比较，如果存在满足要求的，那么ans就是可行的；如果不满足，就求x2x_2x2，这样滑动窗口求下去。

#include <bits/stdc++.h>
#define f(i, n) for(int i = 1; i <= n; i ++)
#define nf(i, n) for(int i = n; i >= 1; i --)
typedef long long ll;
using namespace std;const int N = 2e5 + 10;struct node
{int x, y;
};
node a[N];int mx[N], mn[N];bool cmp(const node &a, const node &b)
{return a.x < b.x;
}int n;bool check(int x)
{int p = 1, q = 1;while (p <= n && q <= n){while (a[q].x - a[p].x < x && q <= n)++ q;if (q > n)return false;if (a[p].y - mn[q] >= x || mx[q] - a[p].y >= x)return true;++ p; }return false;
}int main()
{//freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);cin.tie(0);int T = 1;//cin >> T;while (T --){cin >> n;f(i, n)cin >> a[i].x >> a[i].y;sort (a + 1, a + n + 1, cmp);mx[n] = mn[n] = a[n].y;nf(i, n - 1){mx[i] = max(mx[i + 1], a[i].y);mn[i] = min(mn[i + 1], a[i].y);}int l = 0, r = 1e9, mid, ans;while (l <= r){mid = (l + r) >> 1;if (check(mid)){ans = mid;l = mid + 1;}elser = mid - 1;}cout << ans;}return 0;
}

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