CUGB计算几何专题：B - A Round Peg in a Ground Hole判断线段相交

A - 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.

这题刚开始想的是从后面往回判断相交的，不过……不行，应该是从前往后判断才对，其中的原因可以画图来看……

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define PI acos(-1.0)
#define eps 1e-8
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define f(i,a,n) for(i=a;i<n;i++)
#define F(i,a,n) for(i=a;i<=n;i++)
#define MM 100005
#define MN 505
#define INF 10000007
using namespace std;
typedef long long ll;
inline double sqr(const double &x){ return x * x;}
inline double sgn(const double &x){ return x < -eps ? -1 : x > eps;}
struct Point{
    double x, y;
    Point(const double &x = 0, const double &y = 0):x(x), y(y){}
    Point operator - (const Point &a)const{ return Point(x - a.x, y - a.y);}
    Point operator + (const Point &a)const{ return Point(x + a.x, y + a.y);}
    Point operator * (const double &a)const{ return Point(x * a, y * a);}
    Point operator / (const double &a)const{ return Point(x / a, y / a);}
    bool operator < (const Point &a)const{ return sgn(x - a.x) < 0 || (sgn(x - a.x) == 0 && sgn(y - a.y) < 0);}
    friend double det(const Point &a, const Point &b){ return a.x * b.y - a.y * b.x;}
    friend double dot(const Point &a, const Point &b){ return a.x * b.x + a.y * b.y;}
    friend double dist(const Point &a, const Point &b){ return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}
    double len(){ return sqrt(dot(*this, *this));}

    Point rotateA(const double &angle)const{ return rotateS(cos(angle), sin(angle));}
    Point rotateS(const double &cosa, const double &sina)const{ return Point(x * cosa - y * sina, x * sina + y * cosa);}

    void in(){  scanf("%lf %lf", &x, &y); }
    void out()const{ printf("%.2f %.2f\n",x, y);}
};
struct Line{
    Point s, t;
    Line(const Point &s = Point(), const Point &t = Point()):s(s), t(t){}
    Point dire()const{ return t - s;}
    double len()const{ return dire().len();}
    bool isPointInLine(const Point &p)const{ return sgn(det(p - s, t - s)) == 0 && sgn(dot(p - s, p - t)) <= 0;}
    bool isPointInLineEx(const Point &p)const{ return sgn(det(p - s, t - s)) == 0 && sgn(dot(p - s, p - t)) < 0;}//不含端点
    Point pointProjLine(const Point &p){ return s + dire() * ((dot(p - s, dire()) / dire().len()) /(dire().len()));}
    double pointDistLine(const Point &p){ return fabs(det(p - s, dire()) / dire().len());}

    friend bool sameSide(const Line &line , const Point &a, const Point &b){
        return sgn(det(b - line.s, line.dire())) * sgn(det(a - line.s, line.dire())) > 0;
    }
    friend bool isLineInsectLine(const Line &l1, const Line &l2){
        if(sgn(det(l2.s - l1.s, l1.dire())) == 0 && sgn(det(l2.t - l1.s, l1.dire())) == 0
           && sgn(det(l1.s - l2.s, l2.dire())) == 0 && sgn(det(l1.t - l2.s, l2.dire())) == 0){
            return l1.isPointInLine(l2.s) || l1.isPointInLine(l2.t) || l2.isPointInLine(l1.s) ||l2.isPointInLine(l1.t);
        }
        return !sameSide(l1, l2.s, l2.t) && !sameSide(l2, l1.s, l1.t);
    }

    friend Point lineInsectLine(const Line &l1, const Line &l2){
        double s1 = det(l1.s - l2.s, l2.dire()), s2 = det(l1.t - l2.s, l2.dire());
        return (l1.t * s1 - l1.s * s2) / (s1 - s2);
    }

    void in(){ s.in(); t.in();}
    void out()const{ s.out(); t.out(); }
};
Point a[MM][2];
int vis[MM],s[MM];
int main()
{
    int n;
    while(sca(n)&&n)
    {
        Line b,b1;
        int k=0,i,j;
        mem(vis,0);
        for(i=1;i<=n;i++)
            a[i][0].in(),a[i][1].in();
        for(i=1;i<=n;i++)
        {
            if(vis[i]) continue;
            int flag=0;
            b.s=a[i][0],b.t=a[i][1];
            for(j=i+1;j<=n;j++)
            {
                if(!vis[j])
                {
                    b1.s=a[j][0],b1.t=a[j][1];
                    if(isLineInsectLine(b,b1))
                    {
                        vis[i]=1; flag=1;
                        break;
                    }
                }
            }
            if(!flag) s[k++]=i;
        }
        printf("Top sticks: ");
        for(i=0;i<k;i++)
            if(i!=k-1) printf("%d, ",s[i]);
            else printf("%d.\n",s[i]);
    }
    return 0;
}