题目大意：有$n(n\leqslant300)$个点，每个点坐标范围在$[1\sim100]$，求一个矩阵，使得边界上的点最多。

题解：做一遍二维前缀和，直接暴力枚举两个顶点

卡点：无

 

C++ Code：

#include 
     
      #include 
      
       #define maxn 105int n, ans;int s[maxn][maxn];inline int calc(int lx, int ly, int rx, int ry) {	if (lx > rx || ly > ry) return 0;	return s[rx][ry] - s[rx][ly - 1] - s[lx - 1][ry] + s[lx - 1][ly - 1];}int main() {	scanf("%d", &n);	for (int i = 0, x, y; i < n; ++i) {		scanf("%d%d", &x, &y);		s[x][y] = 1;	}	for (int i = 1; i <= 100; ++i) {		for (int j = 2; j <= 100; j++) s[i][j] += s[i][j - 1];	}	for (int i = 2; i <= 100; ++i) {		for (int j = 1; j <= 100; j++) s[i][j] += s[i - 1][j];	}	for (int lx = 1; lx <= 100; ++lx)		for (int ly = 1; ly <= 100; ++ly)			for (int rx = lx; rx <= 100; ++rx)				for (int ry = ly; ry <= 100; ++ry) {					ans = std::max(ans, calc(lx, ly, rx, ry) - calc(lx + 1, ly + 1, rx - 1, ry - 1));				}	printf("%d\n", ans);	return 0;}