poj 3141Distant Galaxy

Distant Galaxy

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1845 Accepted: 582

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?

Input

There are multiple test cases in the input file. Each test case starts with one integerN, (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 theK^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 Input

102 39 27 43 45 71 510 410 611 44 60

Sample Output

Case 1: 7

Source

Shanghai 2006

 

 

 

#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>using namespace std;const int MAX = 100+5;struct P{     int x,y;     bool operator <(const P& point) const     {            return x < point.x;     }}points[MAX];int caculate(int n){    int ys[MAX];    sort(points, points + n);    for(int i = 0; i < n; i++)ys[i] = points[i].y;        //计算y行数     sort(ys, ys + n);    int y_count = unique(ys, ys + n) - ys;       // for(int i = 0; i < y_count; i++)cout <<" "<<ys[i];    if(y_count < 2)return n;        int left[MAX],vertical[MAX];        // memset(vertical, n, 0);    int ans = 0;    for(int up = 0; up < y_count; up++)        for(int dwn = up + 1; dwn < y_count; dwn++)        {           // int up_axis = y[up],dwn_axis = y[dwn];                        int col = 0;            left[0] = 0,vertical[0] = 0;            for(int i = 0; i < n; i++)            {                if(i != 0 && points[i].x != points[i - 1].x)                {                     vertical[++col] = 0;                     left[col] =  left[col - 1];                 }                                 if(points[i].y == ys[up] || points[i].y == ys[dwn])left[col]++;                if(points[i].y > ys[up] && points[i].y < ys[dwn])vertical[col]++;                            }                        if(!col)return n;            int m = vertical[0];                        //cout << ys[up]<<"---"<<ys[dwn]<<endl;           // cout<<"left-"<<0<<"=>"<<left[0]<<" verti-"<<0<<"=>"<<vertical[0]<<endl;             for(int i = 1; i <= col; i++)            {                                   // cout<<"left-"<<i<<"=>"<<left[i]<<" verti-"<<i<<"=>"<<vertical[i]<<endl;                 ans = max(ans, left[i] + vertical[i] + m);                m = max(m, vertical[i] - left[i - 1]);                            }                    }    return ans;}int main(){        int  t,n,x,y,kase = 1;    //scanf("%d",&t);    while(1)    {        scanf("%d",&n);        if(!n)break;        for(int i = 0; i < n; i++)            scanf("%d%d",&points[i].x, &points[i].y);                int ans = caculate(n);           cout <<"Case "<<kase++<<": "<< ans<<endl;    }    //system("pause");    return 0;}