poj 2606_Rabbit hunt_计算几何

题目描述

给你n个点，求最多有多少点共线

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思路

暴力O(n^3)枚举每个点，然后看是否在同一直线上

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#include <stdio.h>using namespace std;inline int read(){    int x=0,p=1;char ch=getchar();    while (ch<'0'||ch>'9'){if (ch=='-')p=-1;ch=getchar();}    while (ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}    return x*p;}struct arr{    double x,y;    void intput(){x=read();y=read();}}edge[400];arr r;int main(){    int n;    scanf("%d",&n);    for (int i=1;i<=n;i++)        edge[i].intput();       int ans=0;    for (int i=1;i<=n;i++)    {        for (int j=i+1;j<=n;j++)        {            int t=0;            for (int k=j+1;k<=n;k++)                if ((edge[j].x-edge[i].x)*(edge[k].y-edge[j].y)==(edge[j].y-edge[i].y)*(edge[k].x-edge[j].x))                    t++;            if (ans<t) ans=t;        }    }    printf("%d\n",ans+2);}