题目大意：给出 n 个点，问说一个平行与 x 轴和 y 轴的矩形，最多能有多少个点落在边上。

解题思路：首先先将 y 轴相同的放在一起，然后枚举左右边界，考虑上下边界，维护最大值

#include 
   
     #include 
    
      using namespace std; struct Point { int x; int y; bool operator < (const Point& a) const { return x < a.x; } }; const int maxn = 100 + 10; Point P[maxn]; int y[maxn], on[maxn], on2[maxn], left[maxn]; int solve(int N) { sort(P, P + N); sort(y, y + N); int m = unique(y, y + N) - y; if (m <= 2) return N; int ans = 0; for (int a = 0; a < m; a++) for (int b = a + 1; b < m; b++) { int ymin = y[a], ymax = y[b]; int k = 0; for (int i = 0; i < N; i++) { if (i == 0 || P[i].x != P[i-1].x) { k++; on[k] = on2[k] = 0; left[k] = left[k-1] + on2[k-1] - on[k-1]; } if (P[i].y > ymin && P[i].y < ymax) on[k]++; if (P[i].y >= ymin && P[i].y <= ymax) on2[k]++; } if (k <= 2) return N; int M = 0; for (int j = 1; j <= k; j++) { ans = max(ans, left[j] + on2[j] + M); M = max(M, on[j] - left[j]); } } return ans; } int main() { int N, kase = 0; while (scanf(%d, &N), N) { for (int i = 0; i < N; i++) { scanf(%d%d, &P[i].x, &P[i].y); y[i] = P[i].y; } printf(Case %d: %d , ++kase, solve(N)); } return 0; } 
    
   

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