Ultraviolet-1382 Distant Galaxy

Ultraviolet-1382 Distant Galaxy

N vertices are given, and a rectangle parallel to the x and y axes can have at most many vertices on the edge.

Solution: first, put the y axis together, then enumerate the left and right borders, consider the upper and lower boundary, and maintain the maximum value.

#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;}