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 <algorithm>
#include <cstdio>
#define maxn 105
int 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;
}
原文：https://www.cnblogs.com/Memory-of-winter/p/10336007.html