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=,double yy=)
     {
         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=,w=; char c=getchar();
     while (!isdigit(c)) {if (c=='-') w=-; c=getchar();}
     while (isdigit(c)) x=x*+c-'', 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=; i<=n; ++i)
     {
         k=read(); b=read();
         L[i]=(Line){k,b};
         L[i].x=(Point){,b};
         L[i].y=(Point){,k+b};
         L[i].id=i;
     }
     sort(L+,L+n+);
     for (int i=; i<=n; ++i)
     {
         if (top && L[i].k==stack[top].k) continue;
         while (top>=)
         {
             Point p=Line_Cross(stack[top],L[i]);
             if (Cross(stack[top-].x-p,stack[top-].y-p)>) break;
             top--;
         }
         stack[++top]=L[i];
     }
     for (int i=; i<=top; ++i) ans[stack[i].id]=;
     for (int i=; i<=n; ++i) if (ans[i]) printf("%d ",i);
 }