BZOJ1007:[HNOI2008]水平可见直线(计算几何)

Description

在xoy直角坐标平面上有n条直线L1,L2,...Ln,若在y值为正无穷大处往下看,能见到Li的某个子线段,则称Li为
可见的,否则Li为被覆盖的.
例如,对于直线:
L1:y=x; L2:y=-x; L3:y=0
则L1和L2是可见的,L3是被覆盖的.
给出n条直线,表示成y=Ax+B的形式(|A|,|B|<=500000),且n条直线两两不重合.求出所有可见的直线.

Input

第一行为N(0 < N < 50000),接下来的N行输入Ai,Bi

Output

从小到大输出可见直线的编号，两两中间用空格隔开,最后一个数字后面也必须有个空格

Sample Input

3
-1 0
1 0
0 0

Sample Output

1 2

Solution

首先把直线按照斜率排序，再用个栈维护一下。

画个图可以发现，如果直线$i$和直线$stack[top]$的交点在直线$stack[top-1]$的左边，那么$stack[top]$就可以被弹出了。随便判判就好了。

Code

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define N (50009)
#define INF (1e18)
using namespace std;

struct Vector
{
    double x,y;
    Vector(double xx=0,double yy=0)
    {
        x=xx; y=yy;
    }
}p[N];
typedef Vector Point;

struct Line
{
    double k,b;
    Point x,y;
    int id;
    bool operator < (const Line &a) const
    {
        return k==a.k?b>a.b:k>a.k;
    }
    bool operator == (const Line &a) const
    {
        return k==a.k && b==a.b;
    }
}L[N],stack[N];

int n,k,b,ans[N],top;

double Cross(Vector a,Vector b) {return a.x*b.y-a.y*b.x;}
Vector operator - (Vector a,Vector b) {return Vector(a.x-b.x,a.y-b.y);}

inline int read()
{
    int x=0,w=1; char c=getchar();
    while (!isdigit(c)) {if (c=='-') w=-1; c=getchar();}
    while (isdigit(c)) x=x*10+c-'0', c=getchar();
    return x*w;
}

Point Line_Cross(Line u,Line v)
{
    Point ans;
    ans.x=(v.b-u.b)/(u.k-v.k);
    ans.y=u.k*ans.x+u.b;
    return ans;
}

int main()
{
    n=read();
    for (int i=1; i<=n; ++i)
    {
        k=read(); b=read();
        L[i]=(Line){k,b};
        L[i].x=(Point){0,b};
        L[i].y=(Point){1,k+b};
        L[i].id=i;
    }
    sort(L+1,L+n+1);
    for (int i=1; i<=n; ++i)
    {
        if (top && L[i].k==stack[top].k) continue;
        while (top>=2)
        {
            Point p=Line_Cross(stack[top],L[i]);
            if (Cross(stack[top-1].x-p,stack[top-1].y-p)>0) break;
            top--;
        }
        stack[++top]=L[i];
    }
    for (int i=1; i<=top; ++i) ans[stack[i].id]=1;
    for (int i=1; i<=n; ++i) if (ans[i]) printf("%d ",i);
}