POJ2653

 Pick-up sticks

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit   Status

Description

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 for this case. The following n lines contain four numbers each, these numbers are the planar coordinates of the endpoints of one stick. The sticks are listed in the order in which Stan has thrown them. You may assume that there are no more than 1000 top sticks. The input is ended by the case with n=0. This case should not be processed.

Output

For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown.  

The picture to the right below illustrates the first case from input.

Sample Input

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.

Hint

Huge input,scanf is recommended.

题目大意：按照一定的顺序扔小木棍，要找出最上层的木棍。

思路：我们可以从下向上找，如果一个线段的上面有与它交叉的线段，就说明这个木棍不合要求，上面一定有其他木棍.

那么问题关键就是线段交叉了。 都知道2个点在直线两边的条件是叉积的符号相反。

那么我们判断2次，即两个木棍的两个端点都在另一个木棍的两边，就说明线段交叉了。

#include <cstdlib>
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const double eps = 1E-8;
struct point
{
       double x,y;
};
struct line
{
       point a,b;
       int c;
};

int sig(double d)
{
       return (d>eps)-(d<-eps);
}
double cross(point &o,point &a,point &b)
{
       return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
line l[100003];
int main()
{
    int i,j,n;
    while(scanf("%d",&n),n)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%lf %lf %lf %lf",&l[i].a.x,&l[i].a.y,&l[i].b.x,&l[i].b.y);
            l[i].c=1;
        }
        for(i=1;i<n;i++)
        {
           for(j=i+1;j<=n;j++)
           {
               int d1,d2,d3,d4;
               d1=sig(cross(l[i].a,l[j].a,l[j].b));
               d2=sig(cross(l[i].b,l[j].a,l[j].b));
               d3=sig(cross(l[j].a,l[i].a,l[i].b));
               d4=sig(cross(l[j].b,l[i].a,l[i].b));
               if((d1^d2)==-2&&(d3^d4)==-2)
               {
    //   printf("[%d %d %d %d]\n",d1,d2,d3,d4);
                   l[i].c=0;
                   break;
               }
           }
        }
        printf("Top sticks: ");
        for(i=1;i<n;i++)
        if(l[i].c)
        printf("%d, ",i);
        printf("%d.\n",n);
    }
    system("PAUSE");
    return EXIT_SUCCESS;
}