poj 2653 Pick-up sticks(判断线段相交)

题意:在桌上一次放n根木棒,求最上面的木棒编号;

思路:暴力枚举,看每根木棒上是否有木棒;

技巧:使用叉积,判断一个向量的两端点是否在另一向量的同侧,从而判断相交;

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const double epsi=1e-10;
const int maxn=100005;
inline int sign(const double &x){
    if(x>epsi) return 1;
    if(x<-epsi) return -1;
    return 0;
}
struct point{
    double x,y;
    point(){}
    point(double xx,double yy):x(xx),y(yy){}
    point operator -(const point &op2) const{
        return point(x-op2.x,y-op2.y);
    }
    double operator ^(const point &op2) const{
        return x*op2.y-y*op2.x;
    }
};
inline double sqr(const double &x){
    return x*x;
}
inline double mul(const point &p0,const point &p1,const point &p2){
    return (p1-p0)^(p2-p1);
}
inline double dis2(const point &p1,const point &p2){
    return sqr(p1.x-p2.x)+sqr(p1.y-p2.y);
}
inline double dis(const point &p0,const point &p1){
    return sqrt(dis2(p0,p1));
}
inline int cross(const point &p1,const point &p2,const point &p3,const point &p4)
{
    double a1=mul(p1,p2,p3),a2=mul(p1,p2,p4);
    if(sign(a1)==0&&sign(a2)==0) return 2;  //两向量共线
    if(sign(a1)==sign(a2)) return 0;  //两向量不相交
    return 1;       //两向量相交
}
int n;
point p1[maxn],p2[maxn];
int main()
{
    int t,i,j,k;
    while(scanf("%d",&n)){
        if(n==0) break;

        int f1=0;
        for(i=1;i<=n;i++)
        {
            scanf("%lf %lf %lf %lf",&p1[i].x,&p1[i].y,&p2[i].x,&p2[i].y);
        }
        printf("Top sticks:");
        for(i=1;i<=n;i++)
        {
            int flag=0;
            for(j=i+1;j<=n;j++)
            {
                if(cross(p1[i],p2[i],p1[j],p2[j])&&cross(p1[j],p2[j],p1[i],p2[i]))//因为是线段,所以要相互判断
                {
                    flag=1;break;
                }
            }
            if(flag==0&&f1) printf(",");
            if(flag==0) printf(" %d",i),f1=1;
        }
        printf(".\n");
    }
    return 0;
}

 

posted on 2015-05-29 23:48  大树置林  阅读(143)  评论(0编辑  收藏  举报

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