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-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;

 struct point
 {
     int x,y;
     point(int x=,int y=):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<?-:;
 }

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

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