//其实刚遇到这道题目的时候是会做的,,但是却找不到线段相交的模板实在让我无语。。
//意识到模板的重要性了。。
/***********************************
暴力就行,从第一个开始判断
如果两条线段相交就把前面一条筛选掉
判断线段相交直接贴的吉大模板。。。
***********************************/
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn=100000+5;
const double eps=1e-10;

struct point
{
 double x,y;
};

point p[maxn],b[maxn];
bool ans[maxn];

double min(double a,double b)
{
 return a< b?a:b;
}

double max(double a, double b)
{
 return a>b?a:b;
}

bool inter(point a,point b,point c,point d)
{
    if( min(a.x, b.x)>max(c.x,d.x)||
        min(a.y, b.y)>max(c.y,d.y)||
        min(c.x, d.x)>max(a.x,b.x)||
        min(c.y, d.y)>max(a.y, b.y) )
    return 0;

    double h,i,j,k;

    h=(b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
    i=(b.x-a.x)*(d.y-a.y)-(b.y-a.y)*(d.x-a.x);
    j=(d.x-c.x)*(a.y-c.y)-(d.y-c.y)*(a.x-c.x);
    k=(d.x-c.x)*(b.y-c.y)-(d.y-c.y)*(b.x-c.x);

    return h*i<=eps&&j*k<=eps;
}

int main()
{
    int n,i,j;
    int res[maxn];
    while(cin>>n,n)
    {
        memset(ans,0,sizeof(ans));
        for(i=0; i<n; i++ )
        {
            cin>>p[i].x>>p[i].y>>b[i].x>>b[i].y;
        }

        for(i=0; i<n; i++ )
        {
            for(j=i+1;j<n;j++)
            {
                if(inter(p[i],b[i],p[j],b[j]) )
                {
                    ans[i]=1;
                    break;              //不加break会超时。。。
                }
            }
        }
        int ct=0;
        cout<< "Top sticks: ";
        for(i=0; i<n; i++ )
            if(!ans[i])
    res[ct++]=i+1;
        for(i=0; i<ct-1; i++ )
            cout<<res[i]<<", ";
        cout<<res[ct-1]<<"."<<endl;
    }
    return 0;
}

posted on 2011-08-28 00:36  →木头←  阅读(332)  评论(0编辑  收藏  举报