HDU 1432 Lining Up （POJ 1118）

枚举，枚举点 复杂度为n^3。

还可以枚举边的，n*n*log(n)。

POJ 1118 要判断0退出。

#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<stack>
#include<iostream>
#include<list>
#include<set>
#include<vector>
#include<cmath>

#define INF 0x7fffffff
#define eps 1e-8
#define LL long long
#define PI 3.141592654
#define CLR(a,b) memset(a,b,sizeof(a))
#define FOR(i,a,n) for(int i= a;i< n ;i++)
#define FOR0(i,a,b) for(int i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define ft first
#define sd second
#define sf scanf
#define pf printf
#define acfun std::ios::sync_with_stdio(false)

#define SIZE 700+1
using namespace std;

struct lx
{
    int x,y;
}p[SIZE];
int n;

int main()
{
    while(~sf("%d",&n))
    //while(~sf("%d",&n),n)
    {
        FOR(i,0,n)
        sf("%d%d",&p[i].x,&p[i].y);
        int ans=0;
        int maxn=0;
        FOR(i,0,n)
        {
            FOR(j,i+1,n)
            {
                maxn=0;
                FOR(k,j+1,n)
                {
                    if((p[j].x-p[i].x)*(p[k].y-p[j].y)==(p[j].y-p[i].y)*(p[k].x-p[j].x))
                        maxn++;
                }
                ans=max(maxn,ans);
            }
        }
        pf("%d\n",ans+2);
    }
}