poj 2653 Pick-up sticks 线段相交

最新推荐文章于 2020-06-24 20:37:33 发布

cyendra

最新推荐文章于 2020-06-24 20:37:33 发布

阅读量711

点赞数

CC 4.0 BY-SA版权

分类专栏：解题报告几何文章标签：几何

本文链接：https://blog.youkuaiyun.com/cyendra/article/details/9105845

解题报告同时被 2 个专栏收录

260 篇文章

订阅专栏

几何

8 篇文章

订阅专栏

本文介绍了一个算法问题，即如何找出所有顶部的木棍。输入包括一系列木棍的坐标，输出则是那些没有其他木棍覆盖的顶部木棍的列表。文章提供了一种通过判断线段是否相交的方法来解决该问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

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

Hint

Huge input,scanf is recommended.

Source

Waterloo local 2005.09.17

--------------------

直接判断线段相交即可

--------------------

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<iostream>
#include<sstream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cctype>
#include<string>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<climits>
#include<complex>
#define mp make_pair
#define pb push_back
using namespace std;
const double eps=1e-8;
const double pi=acos(-1.0);
const double inf=1e20;
const int maxp=1111;
int dblcmp(double d)
{
    if (fabs(d)<eps)return 0;
    return d>eps?1:-1;
}
inline double sqr(double x){return x*x;}

struct point
{
    double x,y;
    point(){}
    point(double _x,double _y):
    x(_x),y(_y){};
    void input()
    {
        scanf("%lf%lf",&x,&y);
    }
    void output()
    {
        printf("%.2f %.2f\n",x,y);
    }
    bool operator==(point a)const
    {
        return dblcmp(a.x-x)==0&&dblcmp(a.y-y)==0;
    }
    bool operator<(point a)const
    {
        return dblcmp(a.x-x)==0?dblcmp(y-a.y)<0:x<a.x;
    }
    double len()
    {
        return hypot(x,y);
    }
    double len2()
    {
        return x*x+y*y;
    }
    double distance(point p)
    {
        return hypot(x-p.x,y-p.y);
    }
    point add(point p)
    {
        return point(x+p.x,y+p.y);
    }
    point sub(point p)
    {
        return point(x-p.x,y-p.y);
    }
    point mul(double b)
    {
        return point(x*b,y*b);
    }
    point div(double b)
    {
        return point(x/b,y/b);
    }
    double dot(point p)
    {
        return x*p.x+y*p.y;
    }
    double det(point p)
    {
        return x*p.y-y*p.x;
    }
};
struct line
{
    point a,b;
    line(){}
    line(point _a,point _b)
    {
        a=_a;
        b=_b;
    }
    bool operator==(line v)
    {
        return (a==v.a)&&(b==v.b);
    }
    void input()
    {
        a.input();
        b.input();
    }
    double length()
    {
        return a.distance(b);
    }
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int segcrossseg(line v)
    {
        int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
        int d3=dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
        int d4=dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
        if ((d1^d2)==-2&&(d3^d4)==-2)return 2;
        return (d1==0&&dblcmp(v.a.sub(a).dot(v.a.sub(b)))<=0||
                d2==0&&dblcmp(v.b.sub(a).dot(v.b.sub(b)))<=0||
                d3==0&&dblcmp(a.sub(v.a).dot(a.sub(v.b)))<=0||
                d4==0&&dblcmp(b.sub(v.a).dot(b.sub(v.b)))<=0);
    }
};
int n;
line a[110000];
bool v[110000];

int main()
{
    while (~scanf("%d",&n))
    {
        if (n==0) break;
        for (int i=1;i<=n;i++)
        {
            a[i].input();
        }
        for (int i=1;i<=n;i++)
        {
            for (int j=i+1;j<=n;j++)
            {
                if (a[i].segcrossseg(a[j])!=0)
                {
                    v[i]=true;
                    break;
                }
            }
        }
        printf("Top sticks:");
        int num=0;
        for (int i=1;i<=n;i++)
        {
            if (!v[i])
            {
                if (!num) printf(" %d",i);
                else printf(", %d",i);
                num++;
            }
            else v[i]=false;
        }
        printf(".\n");

    }
}