BZOJ1173 Balkan2007 Point

提示：
1. 如何判断三个点共线 ， 跟二维的情形是一样的啊
2. 去重的话各种yy方法都可以啦

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cassert>
using namespace std;
const int maxn = 1100;
int g , t;

struct points
{
    int x , y , z , re;
    points(){}
    void read() { scanf("%d%d%d",&x,&y,&z); }
    points(int _x,int _y, int _z):x(_x),y(_y),z(_z)
    {
        if(x) t = x; else if(y) t = y; else t = z; assert(t);

        if(t<0) x*=-1 , y*=-1 , z*=-1 , re = 1;
        else re = 0;

        g = t;
        if(x) g = __gcd(g , x); if(y) g = __gcd(g , y); if(z) g = __gcd(g , z);
        g = abs(g);
        x/=g; y/=g; z/=g;
    }

    bool operator <(const points& b)const 
    {
        return x<b.x || (x==b.x && y<b.y) || (x==b.x && y==b.y && z<b.z);
    }
    bool operator ==(const points& b)const 
    {
        return x==b.x && y==b.y && z==b.z;
    }
};

int n , res;
points a[maxn] , b[maxn];

__inline void solve(int cnt)
{
    bool ok;
    sort(b, b+cnt);
    for(int i=0 , j;i<cnt;i = j)
    {
        for(j=i;j<cnt && b[j]==b[i];j++) ;
        if(j-i>=2)
        {
            ok = true;
            for(int k=i;k<j;k++) if(b[k].re) { ok = false; break; }
            res+= ok;
        }
    }
}

int main(int argc, char *argv[]) {

    scanf("%d" , &n);
    for(int i=1;i<=n;i++) a[i].read();

    for(int i=1;i<=n;i++)
    {
        int cnt = 0;
        for(int j=1;j<=n;j++) if(j!=i) b[cnt++] = points(a[j].x-a[i].x , a[j].y-a[i].y , a[j].z-a[i].z);
        solve(cnt);
    }

    cout<<res<<endl;
    return 0;
}