POJ 1118 Lining Up 直线穿过最多的点数

http://poj.org/problem?id=1118

直接枚举O(n^3) 1500ms能过...数据太水了...这个代码就不贴了...

斜率排序O(n^2logn)是更好的做法...枚举斜率...直线方程相同的线段数量k...一定满足方程 n(n-1)/2=k  ---------------  还真是baka. 到2点才想出这个结论

特判只有一个点,两个点的情况...500ms AC

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;

#define EPS         1e-8
#define MAXN        (int)5e5+5
#define MOD         (int)1e9+7
#define PI          acos(-1.0)
#define LINF        ((1LL)<<50)
#define INF         (1<<30)
#define DINF        (1e10)
#define max(a,b)    ((a) > (b) ? (a) : (b))
#define min(a,b)    ((a) < (b) ? (a) : (b))
#define max3(a,b,c) (max(max(a,b),c))
#define min3(a,b,c) (min(min(a,b),c))
#define BUG         cout<<"BUG! "<<endl
#define LLL         cout<<"--------------"<<endl
#define L(t)        (t << 1)
#define R(t)        (t << 1 | 1)
#define Mid(a,b)    ((a + b) >> 1)
#define lowbit(a)   (a & -a)
#define FIN         freopen("in.txt","r",stdin)
#define FOUT        freopen("out.txt","w",stdout)
#pragma comment     (linker,"/STACK:102400000,102400000")

// typedef long long LL;
// typedef unsigned long long ULL;
// typedef __int64 LL;
// typedef unisigned __int64 ULL;
// int gcd(int a,int b){ return b?gcd(b,a%b):a; }
// int lcm(int a,int b){ return a*b/gcd(a,b); }

/*********************   F   ************************/
struct POINT{
    double x,y;
    POINT(double _x = 0, double _y = 0):x(_x),y(_y){};
    int id;
}p[800];
struct LINE{
    POINT a,b;
    double K,B;
    LINE(POINT _a = 0,POINT _b = 0):a(_a),b(_b){
        if((a.x - b.x) == 0){
            K = DINF;
            B = a.x;
        }else{
            K = ((a.y - b.y) / (a.x - b.x));
            B = (- (b.x * a.y - a.x * b.y) / (a.x - b.x));
        }
    }
}L[500000];
bool cmp(LINE a,LINE b){
    if(a.K == b.K) return a.B < b.B;
    return a.K < b.K;
}
int main()
{
    //FIN;
    int n;
    while(cin>>n && n){
        for(int i = 0 ; i < n ; i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        if(n == 1 || n == 2){
            printf("%d
",n);
            continue;
        }
        int cnt = 0;
        for(int i = 0 ; i < n ; i++)
            for(int j = i+1 ; j < n ; j++)
                L[cnt++] = LINE(p[i],p[j]);
        sort(L,L+cnt,cmp);
        int m = -INF;
        for(int i = 0 ; i < cnt-1 ; i++){
            int c = 0;
            while(L[i].K == L[i+1].K && L[i].B == L[i+1].B){
                c++;
                i++;
            }
            c = sqrt((c+1)*2)+1;
            m = max(m,c);
        }
        printf("%d
",m);
    }
    return 0;
}