POJ 2007 简单几何

既然是整数,就不要写double了.

既然是整数,就用叉积吧,别用atan2 丢失精度.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;
const int maxn = 50+3;

struct Point
{
    int x;
    int y;
}p[maxn],P0;

int Mult(Point p0, Point p1, Point p2)  //叉积
{
    return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y);
}

int cmp(const Point &p1, const Point &p2)  //排序函数
{
    return Mult(P0,p1,p2)>0;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
#endif
    int n = 0;
    while(cin>>p[n].x>>p[n].y)
    {
        n++;
    }
    P0 = p[0];
    sort(p+1,p+n,cmp);
    for(int i = 0; i < n; i++)
    {
        cout<<"("<<p[i].x<<","<<p[i].y<<")"<<endl;
    }
    return 0;
}

我试着写了下用double的(仅仅求角度换成了double 比较斜率大小 用了精度判断(<1e-8) ) 依然WA

无解...这种题目还是老老实实 写整数的吧.多好...能不丢失精度就不丢失精度~