UVALive - 3695:Distant Galaxy

Distant Galaxy
出典:UVALLive
ラベル:
参考資料:『アルゴリズムコンテスト入門経典——トレーニングガイド』P 52
類似タイトル:https://blog.csdn.net/wingrez/article/details/82426599
タイトル
You are observing a distant galaxy using a telescope above the Astronomy Tower, and you think that a rectangle drawn in that galaxy whose edges are parallel to coordinate axes and contain maximum star systems on its edges has a great deal to do with the mysteries of universe. However you do not have the laptop with you, thus you have written the coordinates of all star systems down on a piece of paper and decide to work out the result later. Can you finish this task?
入力
There are multiple test cases in the input file. Each test case starts with one integer N, (1 ≤ N ≤ 100), the number of star systems on the telescope. N lines follow, each line consists of two integers: the X and Y coordinates of the K-th planet system. The absolute value of any coordinate is no more than 109, and you can assume that the planets are arbitrarily distributed in the universe. N = 0 indicates the end of input file and should not be processed by your program.
しゅつりょく
For each test case, output the maximum value you have found on a single line in the format as indicated in the sample output.
入力サンプル
10 2 3 9 2 7 4 3 4 5 7 1 5 10 4 10 6 11 4 4 6 0
出力サンプル
Case 1: 7
リファレンスコード

#include
#include
#define MAXN 105
using namespace std;

struct Point{
	int x,y;
	bool operator < (const Point& rhs) const{
		return x<rhs.x; 
	}
}P[MAXN];

int n;
int y[MAXN];
int on[MAXN],on2[MAXN],left[MAXN];

int solve(){
	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]= k==0?0: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 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());
	}
	return 0;
}