Codeforces Round #276 (Div. 2) B. Valuable Resources 二分

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由于边界与坐标轴平行，则只需确定正方形的左下角和边长即可确定正方形。

而左下角可以根据mines的坐标确定，所以只需二分枚举正方形的边长即可，给定边长，O(n)时间内可以检查是否满足。

代码如下：

#include <cstdio>#define N 1005#define INF 1000000001int x[N], y[N], n, orig_x = INF, orig_y = INF;bool check(int len){int right = orig_x + len, up = orig_y + len;for(int i = 0; i < n; i ++){if(!(x[i] >= orig_x && x[i] <= right && y[i] >= orig_y && y[i] <= up))return false;}return true;}int main(){scanf("%d", &n);for(int i = 0; i < n; i ++){scanf("%d %d", &x[i], &y[i]);if(x[i] < orig_x)orig_x = x[i];if(y[i] < orig_y)orig_y = y[i];}int l = 0, r = 2 * INF;while(l < r){int mid = l + ((r - l) >> 1);if(check(mid))r = mid;elsel = mid + 1;}printf("%I64d\n", (long long)l * l);return 0;}

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