POJ1118 Lining up

Lining Up

Time Limit: 2000MS Memory Limit: 32768KTotal Submissions: 19392 Accepted: 6133

Description

"How am I ever going to solve this problem?" said the pilot. 

Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all lie on one line. Can you write a program that calculates this number? 


Your program has to be efficient! 

Input

Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.

Output

output one integer for each input case ,representing the largest number of points that all lie on one line.

Sample Input

51 12 23 39 1010 110

Sample Output

3

Source

East Central North America 1994

给你几个点，求一条直线最多经过几个点。

个人纯暴力。

这题网上貌似说hash可以做，不懂啊，会的大神来给我讲讲啊

#include<iostream>#include<cstdio>#include<cstring>using namespace std;struct node{    double x,y;} p[701];bool v[701][701];int main(){    int n;    while(cin>>n)    {        if(n==0)return 0;        memset(v,0,sizeof(v));        for(int i=0; i<n; i++)scanf("%lf%lf",&p[i].x,&p[i].y);        int a=0,b;        for(int i=0; i<n-1; i++)        {            for(int j=i+1; j<n; j++)//我是以两个点为基点，看还有木有其他点在这两个点所构成的直线上            {                if(v[i][j]==1)continue;//这是优化                b=0;                //double k=(p[i].y-p[j].y)/(p[i].x-p[j].x);                //double c=k*p[j].x-p[j].y;                for(int u=0; u<n; u++)                {                    if((p[u].y-p[i].y)*(p[j].x-p[i].x)==(p[j].y-p[i].y)*(p[u].x-p[i].x))//为了精度准确，尽量少用除号，我因此wa过1次                    {                        b++;                    v[i][u]=1;//优化                    v[u][i]=1;                    v[u][j]=1;                    v[j][u]=1;}                }                if(b>a)a=b;            }        }        cout<<a<<endl;    }    return 0;}