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UVA - 1382 Distant Galaxy

2015-07-13 21:16 417 查看
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 <cstdio>
#include <algorithm>
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\n", ++kase, solve(N));
}
return 0;
}
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