poj2780Linearity(多点共线）

链接

判断最多多少点在一条直线上，

可以枚举每一个点为坐标系的原点，其它点变成相应的位置，然后求得过原点及其点的斜率，排序找一下最多相同的。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define N 1010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;

struct point
{
    int x,y;
    point(int x=0,int y=0):x(x),y(y){}
}p[N];
double o[N];

typedef point pointt;
pointt operator -(point a,point b)
{
    return point(a.x-b.x,a.y-b.y);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return x;
    return x<0?-1:1;
}

int main()
{
    int n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        for(i = 1; i <= n; i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        int maxz=0;
        for(i = 1; i <= n; i++)
        {
            int ans = 0 ,g=0;
            for(j = 1; j <= n ;j++)
            {
                int x = p[j].x-p[i].x;
                int y = p[j].y-p[i].y;
                if(x==0)
                ans++;
                else o[g++] = y*1.0/x;
            }
            sort(o,o+g);
            int num = 2;
            for(j = 1; j < g; j++)
            if(dcmp(o[j]-o[j-1])==0)
            num++;
            else
            {ans = max(ans,num);
                num = 2;

            }
            ans = max(ans,num);
            maxz = max(maxz,ans);
        }
        printf("%d
",maxz);
    }
    return 0;
}