bzoj 1007: [HNOI2008]水平可见直线半平面交

Posted 2020-10-26 lokiii

其实并不算标准半平面交？但是思路差不多
先按照斜率排序，然后用栈维护凸壳，每遇到重斜率或a[i],s[top-1]交点的x轴在s[top],s[top-1]交点左侧，则说明s[top]被a[i],s[top-1]覆盖，弹栈即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=50005;
const double eps=1e-8;
int n,top;
bool v[N];
struct qwe
{
    double a,b;
    int id;
}a[N],s[N];
int sgn(double x)
{
    return x<-eps?-1:x>eps;
}
bool cmp(const qwe &a,const qwe &b)
{
    return sgn(a.a-b.a)==0&&a.b<b.b||a.a<b.a;
}
double jdy(qwe a,qwe b)
{
    return (b.b-a.b)/(a.a-b.a);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lf%lf",&a[i].a,&a[i].b),a[i].id=i;
    sort(a+1,a+1+n,cmp);
    for(int i=1;i<=n;i++)
    {
        // while(top>1&&(sgn(jdy(s[top],s[top-1])-jdy(a[i],s[top-1]))>=0||sgn(s[top].a-a[i].a)==0))
            // top--;
        while(top)
        {
            if(sgn(s[top].a-a[i].a)==0)
                top--;
            else if(top>1&&jdy(a[i],s[top-1])<=jdy(s[top],s[top-1]))
                top--;
            else 
                break;
        }
        s[++top]=a[i];
    }
    for(int i=1;i<=top;i++)
        v[s[i].id]=1;
    for(int i=1;i<=n;i++)
        if(v[i])
            printf("%d ",i);
    return 0;
}