BZOJ 1007 [HNOI2008]水平可见直线 ——半平面交 凸包

发现需要求一个下凸的半平面上有几个交点。

然后我们把它变成凸包的问题。

好写、好调、还没有精度误差。

#include <map>
#include <ctime>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define ll long long
 
struct Vector{
    int x,y;
    void print()
    {
        printf("Vector - > (%d,%d)
",x,y);
    }
};
 
struct Point{
    int x,y;
    int id;
    void print()
    {
        printf("Point ID %d (%d,%d)
",id,x,y);
    }
};
 
Vector operator - (Point a,Point b)
{Vector ret;ret.x=a.x-b.x;ret.y=a.y-b.y;return ret;}
 
ll operator * (Vector a,Vector b)
{return (ll)a.x*b.y-(ll)a.y*b.x;}
 
int n,top=0;
Point a[50005],sta[50005];
 
bool cmp(Point a,Point b)
{return a.x==b.x?a.y<b.y:a.x<b.x;}
bool cmp2(Point a,Point b)
{return a.id<b.id;}
void Andrew()
{
    F(i,1,n)
    if (a[i].x!=a[i-1].x){
        while (top>=2&&(sta[top]-sta[top-1])*(a[i]-sta[top])<=0) top--;
        sta[++top]=a[i];
    }
    sort(sta+1,sta+top+1,cmp2);
    F(i,1,top) printf("%d ",sta[i].id);
    printf("
");
}
 
void Finout()
{
    freopen("bzoj_1007.in","r",stdin);
    freopen("bzoj_1007.out","w",stdout);
}
 
int main()
{
//  Finout(); 
    scanf("%d",&n);
    F(i,1,n)
    {
        scanf("%d%d",&a[i].x,&a[i].y);
        a[i].y=-a[i].y;
        a[i].id=i;
    }
    sort(a+1,a+n+1,cmp);
    Andrew();
}