浙大月赛——ZOJ Problem Set - 3574

直线x=a,x=b

又有n条线段

求吧区域分成几个空间部分

顶点-线段+空间=3

把线段与两条直线相交的点ll,rr求出来

按min(ll,rr)排序判断来降低时间复杂度

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;

struct data
{
    double ll,rr;
    double duan;
    double min;
}s[30099];

double cmp(data a,data b)
{
    return a.min<b.min;
}

int main()
{
    double a,b;
    while(scanf("%lf%lf",&a,&b)!=EOF)
    {
        int n;
        scanf("%d",&n);
        int i,j;
        double k,c,min;
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf",&k,&c);
            s[i].ll=k*a+c;
            s[i].rr=k*b+c;
            s[i].duan=1;
            
            min=s[i].ll;
            if(min>s[i].rr)
                min=s[i].rr;
            s[i].min=min;
        }
        sort(&s[1],&s[n+1],cmp);

        double dian=0,max;
        for(i=1;i<=n;i++)
        {
            max=s[i].ll;
            if(max<s[i].rr)
                max=s[i].rr;

            for(j=i+1;j<=n;j++)
            {
                if((s[j].rr>s[i].rr&&s[j].ll<s[i].ll)||(s[j].rr<s[i].rr&&s[j].ll>s[i].ll))
                {
                    s[j].duan++;
                    s[i].duan++;
                    dian++;
                }
                if(max<=s[j].min)
                    break;
            }
        }

        dian+=n*2;
        double all=0;
        for(i=1;i<=n;i++)
        {
            all+=s[i].duan;
        }
        all+=(n-1)*2;
        
        double temp=3-dian+all;
        printf("%.0lf\n",temp);
    }

    return 0;
}