poj 2653题解

题意:按照顺序丢下n根棍子，求所有未被其他棍子压在下面的棍子(top sticks)序号

1 <= n <= 100000，top sticks数量不超过1000

题解:开一个vector（叫top），每次扔下一根棍子，把它push_back进去(这根棍子当前一定未被压在下面)，然后遍历top，把所有和这个棍子相交的序号都删除掉。

　　理论上时间复杂度可以达到O(n²)，因为虽然答案元素不超过1000，但中间过程vector可以几乎到达n，除非说“任意时刻top sticks数量不超过1000”，但是还是A了，不知道有没有更好的解法。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;

#define rep(i,a,b) for (int i=a;i<=b;++i)

const double eps=1e-10;

struct point{
    double x,y;
    point(){}
    point (double a,double b): x(a),y(b) {}

    friend point operator + (const point &a,const point &b){
        return point(a.x+b.x,a.y+b.y);
    }

    friend point operator - (const point &a,const point &b){
        return point(a.x-b.x,a.y-b.y);
    }

    friend point operator * (const double &r,const point &a){
        return point(r*a.x,r*a.y);
    }

    friend bool operator == (const point &a,const point &b){
        return (abs(a.x-b.x)<eps && abs(a.y-b.y)<eps);
    }

    double norm(){
        return sqrt(x*x+y*y);
    }
};

inline double det(point a,point b) {return a.x*b.y-a.y*b.x;}

inline bool seg_cross_seg(point a,point b,point c,point d)
{
    if (min(c.x,d.x)>max(a.x,b.x) || min(a.x,b.x)>max(c.x,d.x) || min(c.y,d.y)>max(a.y,b.y) || min(a.y,b.y)>max(c.y,d.y))
        return false;
    return det(a-c,d-c)*det(b-c,d-c)<eps && det(c-a,b-a)*det(d-a,b-a)<eps;
}

int n;
point s[200000],t[200000];
vector<int> top;

int main()
{
    bool flag;
    while (~scanf("%d",&n))
    {
        if (n==0) break;
        rep(i,1,n)
            scanf("%lf%lf%lf%lf",&s[i].x,&s[i].y,&t[i].x,&t[i].y);
        top.clear();
        rep(i,1,n)
        {
            for (vector<int>::iterator it=top.begin();it!=top.end();)
            {

                if (seg_cross_seg(s[*it],t[*it],s[i],t[i]))
                {
                    it=top.erase(it);
                }
                else ++it;
            }
            top.push_back(i);
        }
        printf("Top sticks:");
        for(int i=0;i<top.size()-1;++i) printf(" %d,",top[i]);
        printf(" %d.
",top[top.size()-1]);

    }
    return 0;
}