hdu_1147 Pick_up sticks - c++编程基础

Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.

Input Input consists of a number of cases. The data for each case start with 1 ≤ n ≤ 100000, the number of sticks fZ http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"brush:java;">5 1 1 4 2 2 3 3 1 1 -2.0 8 4 1 4 8 2 3 3 6 -2.0 3 0 0 1 1 1 0 2 1 2 0 3 1 0
Sample Output

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.

分析：

解题思路比较简单――每放一根棍子，都判断一下它与它前面的且在顶端的棍子是否相交，相交的话则将相应的棍子从解空间中除去（当前这根暂时是在解空间中的）

但是，如果直接用数组做会超时，然后用链表做（刚开始自己写了个链表，但是写坏掉了，哎，基础啊……），于是最后还是用了list

除数据结构外，最主要的算法就是判断两直线相交了，直接用了模板

代码：

//hdu 1147
#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          using namespace std; const double eps=1e-8; int sgn(double x) { if(fabs(x)
         
          =min(l2.s.x,l2.e.x) && max(l2.s.x,l2.e.x) >=min(l1.s.x,l1.e.x) && max(l1.s.y,l1.e.y) >=min(l2.s.y,l2.e.y) && max(l2.s.y,l2.e.y) >=min(l1.s.y,l1.e.y) && sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e))<=0 && sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e))<=0 ; } int main() { freopen("in.txt","r",stdin); int n; int len; node cur; list
          
            stick; list
           
            ::iterator p; while(scanf("%d",&n),n){ for(int i=1;i<=n;i++){ scanf("%lf%lf%lf%lf",&cur.t.s.x,&cur.t.s.y,&cur.t.e.x,&cur.t.e.y); cur.no=i; stick.push_back(cur); if(i>1){ p=stick.begin(); len=stick.size(); for(int i=1;i