bzoj千题计划176：bzoj1199: [HNOI2005]汤姆的游戏

http://www.lydsy.com/JudgeOnline/problem.php?id=1199

求出圆x的范围

把要判断的点按x从小到大排序

枚举图形

二分出x满足这个图形的一段区间

枚举这段区间内的每个点

圆判断到圆心的距离

矩形判断y

代码不是我的~~~

#include<bits/stdc++.h>
#define N 1000010
using namespace std;
const double eps=1e-7;
struct node
{
    double x1,x2,y1,y2;
    double x,y,r;
    char c;
    bool friend operator < (node a,node b)
    {
        return a.x1<b.x1;
    }
}a[N];
char s[3];
int out[N];
struct point
{
    double x,y;
    int num;
}e[N];
int n,m;
double sqr(double x)
{
    return x*x;
}
bool pd(double x1,double y1,double x,double y,double r)
{
    return sqr(x-x1)+sqr(y-y1)<sqr(r);
}
bool cmp(point a,point b)
{
    return a.x<b.x;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%s",s+1);
        if(s[1]=='r')
        {
            scanf("%lf%lf%lf%lf",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
            if(a[i].x1>a[i].x2) swap(a[i].x1,a[i].x2),swap(a[i].y1,a[i].y2);
            a[i].c=s[1];
        }
        else
        {
            scanf("%lf%lf%lf",&a[i].x,&a[i].y,&a[i].r);
            a[i].x1=a[i].x-a[i].r;a[i].x2=a[i].x+a[i].r;
            a[i].c=s[1];
        }
    }
    for(int i=1;i<=m;i++) scanf("%lf%lf",&e[i].x,&e[i].y),e[i].num=i;
    sort(e+1,e+m+1,cmp);
    for(int i=1;i<=n;i++)
    {
        int ll=-1,rr=-1;
        int l=1,r=m;
        while(l<=r)
        {
            int mid=l+r>>1;
            if(e[mid].x>a[i].x1)
            {
                ll=mid;
                r=mid-1;
            }
            else l=mid+1;
        }
        l=1,r=m;
        while(l<=r)
        {
            int mid=l+r>>1;
            if(e[mid].x<a[i].x2)
            {
                rr=mid;
                l=mid+1;
            }
            else r=mid-1;
        }
        if(ll==-1 || rr==-1) continue;
        for(int j=ll;j<=rr;j++)
        {
            if(a[i].c=='c')
            {
                if(pd(e[j].x,e[j].y,a[i].x,a[i].y,a[i].r)) out[e[j].num]++;
            }
            else
            {
                if(e[j].y<a[i].y2&&e[j].y>a[i].y1) out[e[j].num]++;
            }
        }
    }
    for(int i=1;i<=m;i++)
        cout<<out[i]<<"
";
    return 0;
}