Pick-up sticks POJ

Pick-up sticks

 题目链接：https://vjudge.net/problem/POJ-2653

 题目：

      

 

 思路：暴力判断两线段间是否相交，若相交则去掉前面的线段，将对应棍子的编号输出即可

// 
// Created by HJYL on 2020/1/13.
//
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#define eps 1e-8
#define Inf 0x7fffffff
//#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+100;
struct Point{
    double x,y;
};
double min(double a, double b) { return a < b ? a : b; }

double max(double a, double b) { return a > b ? a : b; }
double Cross(Point &sp, Point &ep, Point &op)
{
    return (sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y);
}
bool IsSegmentIntersect(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 T;
    bool flag[maxn];
    while(~scanf("%d",&T)) {
        memset(flag,0, sizeof(flag));
        if(T==0)
            break;
        Point p[maxn],b[maxn];
        for(int i=0;i<T;i++){
            scanf("%lf%lf%lf%lf", &p[i].x, &p[i].y, &b[i].x, &b[i].y);
        }
        //for(int i=0;i<=pos;i++)
        // cout<<p[i].x<<" "<<p[i].y<<endl;
        for(int i=0;i<T;i++)
        {
            for(int j=i+1;j<T;j++)
            {

                    if(IsSegmentIntersect(p[i],b[i],p[j],b[j]))
                    {
                        flag[i]=1;
                        break;
                    }
            }
        }
        int gg[maxn]={0};
        int re=0;
        for(int i=0;i<T;i++)
            if(!flag[i]) gg[re++]=i+1;
        printf("Top sticks:");
        for(int i=0;i<re-1;i++)
            printf(" %d,",gg[i]);
        printf(" %d.
",gg[re-1]);
    }
    return 0;
}