POJ 3668 分数保存斜率

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一开始的时候直接用 double 来保存斜率，真是 too young too simple .. always naive

2000*2000的数组开不下，只好优化下内存，好在hash这个东西存的时候每个只需要1 bit ，一个int就能装32bit拉。

最后斜率为0和斜率为无穷大要特判一下。

#include <stdio.h>#include <iostream>#include <queue>#include <algorithm>#include <map>#include <vector>#include <cmath>#include <string.h>#include <stdlib.h>#include <time.h>#include <fstream>#include <set>#include <stack>using namespace std;#define READ freopen("acm.in","r",stdin)#define WRITE freopen("acm.out","w",stdout)#define ll long long#define ull unsigned long long #define PII pair<int,int>#define PDI pair<double,int>#define PDD pair<double,double>#define MII map<int,int>::iterator #define fst first#define sec second#define MS(x,d) memset(x,d,sizeof(x))#define INF 0x3f3f3f3f#define ALL(x) x.begin(),x.end()#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ROOT 0,n-1,1#define PB push_back#define FOR(a,b,c) for(int a=b;a<c;a++)#define MOD 1000000007#define keyTree (ch[ ch[root][1] ][0])#define MAX 200000int n;int x[500],y[500];int hash[2][2100][2100/32];void setbit(int x,int y,int pos){    hash[pos][x][y/32]|=(1<<(y%32));}bool getbit(int x,int y,int pos){    return hash[pos][x][y/32]&(1<<(y%32));}int main(){    //READ;    while(cin>>n)    {        MS(hash,0);        for(int i=0;i<n;i++)            scanf("%d%d",&x[i],&y[i]);        int ans=0;        int flag=0;        for(int i=0;i<n;i++)            for(int j=i+1;j<n;j++)            {                if(x[j]==x[i])                    flag|=1;                else if(y[j]==y[i])                    flag|=2;                else                {                    int dx=x[j]-x[i];                    int dy=y[j]-y[i];                    int pos=dy*dx>0?1:0;                    dx=dx>0?dx:-dx;                    dy=dy>0?dy:-dy;                    int g=__gcd(dx,dy);                    dx/=g,dy/=g;                    if(!getbit(dx,dy,pos))                    {                        ans++;                        setbit(dx,dy,pos);                    }                }            }           if(flag>0&&flag<3)            ans++;        else if(flag==3)            ans+=2;        cout<<ans<<endl;    }    return 0;}

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