Distant Galaxy

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Description

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?

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Input

There are multiple test cases in the input file. Each test case starts with one integer N , (1 $\le$ N $\le$ 100) , the number of star systems on the telescope. N lines follow, each line consists of two integers: the X and Ycoordinates 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.

Output

For each test case, output the maximum value you have found on a single line in the format as indicated in the sample output.

Sample Output

10 2 3 9 2 7 4 3 4 5 7 1 5 10 4 10 6 11 4 4 6 0

Sample Input

Case 1: 7

这道题与最大子矩阵的题有些相似，仍然采取扫描法。先确定矩形的上下界，然后再用一条线从左到右扫描，每扫描完一次求一下能包含的最多的点的个数，，求个数时用维护最大值的方法做。这道题还用到了unique去重函数，该函数是对相邻的数进行去重，一般要先排好顺序，同时该函数并不是把重复的数删掉而是把它们挪到后面，return：An iterator to the element that follows the last element not removed

代码如下：

#include<cstdio>#include<algorithm>#include<iostream>using namespace std;int Y[103],on[103],on2[103],Left[103],N;struct Node{int x,y;bool operator<(const Node &a)const{return x<a.x;}}zb[103];int solve(){int m,ans=0;sort(zb,zb+N);sort(Y,Y+N);m=unique(Y,Y+N)-Y;if(m<=2) return N;for(int i=0;i<m;i++){for(int j=i+1;j<m;j++){int ymin=Y[i],ymax=Y[j],k=0;Left[k]=0,on[k]=0,on2[k]=0;if(zb[k].y>ymin&&zb[k].y<ymax) on[k]++;if(zb[k].y>=ymin&&zb[k].y<=ymax) on2[k]++;for(int t=1;t<N;t++){if(zb[t].x!=zb[t-1].x){k++;on[k]=0;on2[k]=0;        //每扫描到新的位置，一定要清零Left[k]=Left[k-1]+on2[k-1]-on[k-1];}if(zb[t].y>ymin&&zb[t].y<ymax) on[k]++;if(zb[t].y>=ymin&&zb[t].y<=ymax) on2[k]++;}if(k<=1) return N;int s=0;for(int t=0;t<=k;t++){ans=max(ans,Left[t]+on2[t]+s);s=max(s,on[t]-Left[t]);}}}return ans;}int main(){int count=0;while(scanf("%d",&N)==1&&N){int x,y;for(int i=0;i<N;i++){cin>>zb[i].x>>zb[i].y;Y[i]=zb[i].y;}printf("Case %d: %d\n",++count,solve());}return 0;}

这道题我犯了两个错误：

1.数组on【】，on2【】没有赋初值就直接进行加一操作。

2.输出格式有错误。

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