bzoj 4814: [Cqoi2017]小Q的草稿【计算几何】

//先打个50暴力,10min50分简直美滋滋~
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=5005;
const double eps=1e-8;
int n,t,m,ans;
struct dian
{
    double x,y;
    dian(double X=0,double Y=0)
    {
        x=X,y=Y;
    }
    dian operator + (const dian &a) const
    {
        return dian(x+a.x,y+a.y);
    }
    dian operator - (const dian &a) const
    {
        return dian(x-a.x,y-a.y);
    }
}a,b,c,p[N];
struct bian
{
    dian s,t;
    bian(dian S=dian(),dian T=dian())
    {
        s=S,t=T;
    }
}l[N];
int read()
{
    int r=0,f=1;
    char p=getchar();
    while(p>‘9‘||p<‘0‘)
    {
        if(p==‘-‘)
            f=-1;
        p=getchar();
    }
    while(p>=‘0‘&&p<=‘9‘)
    {
        r=r*10+p-48;
        p=getchar();
    }
    return r*f;
}
int sgn(double x)
{
    return x<-eps?-1:x>eps;
}
double cj(dian a,dian b)
{
    return a.x*b.y-a.y*b.x;
}
bool lc(dian a,dian b,dian c,dian d)
{
    return sgn(cj(c-a,b-a)*cj(b-a,d-a))>=0;
}
bool jiao(dian a,dian b,dian c,dian d)
{
    return lc(a,b,c,d)&&lc(c,d,a,b);
}
bool ok(dian a,dian b)
{
    for(int i=1;i<=m;i++)
        if(jiao(a,b,l[i].s,l[i].t))
            return 0;
    return 1;
}
int main()
{
    n=read(),t=read();
    for(int i=1;i<=n;i++)
        p[i].x=read(),p[i].y=read();
    for(int i=1;i<=t;i++)
    {
        a.x=read(),a.y=read(),b.x=read(),b.y=read(),c.x=read(),c.y=read();
        l[++m]=bian(a,b),l[++m]=bian(b,c),l[++m]=bian(c,a);
    }
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
            if(ok(p[i],p[j]))
                ans++;
    printf("%d\n",ans);
    return 0;
}

原文地址：https://www.cnblogs.com/lokiii/p/8511628.html