成都预赛 Polar intersecting line segments 已知N个线段，判断是否相交 排序+划分区间

Polar intersecting line segments

Time Limit: 1000 MS Memory Limit: 65536 K

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Description

In this problem, you are asked to determine if a set of line segments intersect.

Input

The first line of input is a number c ≤ 20, the number of test cases. The first line of each test case is a number k ≤ 50000, the number of line segments. The following k lines each consist of two numbers to eight decimal places θ₁, θ₂ with 0 < θ₁ < 2π and 0 < θ₂ < 2π describing the line segment from (cos θ₁, sin θ₁) to (cos θ₂, sin θ₂). For each test case, all inputs points are distinct.

Output

For each test case, you should output a single line containing “intersect” if any two of the k lines intersect and “do not intersect” otherwise. The endpoints of a line are considered to be part of the line.

Sample Input

2
2
0 3.14159265
1.57079633 4.71238898
2
0 1.57079633
3.14159265 4.71238898

Sample Output

intersect
do not intersect

Source

Northeast North-America 2009, practice contest

 

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e-8;
struct Point
{
    double x,y;//角度
};
Point a[50000+10];
int flag;
bool cmp(Point h,Point k)
{
    if(h.x!=k.x) return h.x<k.x;
    return h.y<k.y;
}
void check(int l,int r)
{
    if(l>=r) return ;
    if(flag==0) return ;
    for(int i=l;i<=r;)
    {
        int j=i+1;
        while(j<=r&&a[i].x<a[j].x&&a[j].y<a[i].y) j++;
        check(i+1,j-1);
        if(flag==0) return ;
        if(j==r+1) return ;
        if(a[i].y<a[j].x) i=j;
        else
        {
            flag=0;
            return ;
        }
    }
}
int main()
{
    int ci;scanf("%d",&ci);
    while(ci--)
    {
        int n;scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&a[i].x,&a[i].y);
            if(a[i].x>a[i].y) swap(a[i].x,a[i].y);
        }
        sort(a,a+n,cmp);
        flag=1;
        check(0,n-1);
        if(flag) printf("do not intersect/n");
        else printf("intersect/n");
    }
    return 0;
}