BZOJ 1818: [Cqoi2010]内部白点 扫描线+树状数组

问题转化为求每一个极长横线段与极长纵线段的交点个数.

这个东西用扫描线+树状数组维护一下就可以了.

code:

#include <cstdio>
#include <algorithm>
#define N 200005
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
namespace BIT
{
    int C[N];
    int lowbit(int t) { return t&(-t); }
    void update(int x,int v)
    {
        for(;x<N;x+=lowbit(x)) C[x]+=v;
    }
    int query(int x)
    {
        int tmp=0;
        for(;x>0;x-=lowbit(x)) tmp+=C[x];
        return tmp;
    }
};
struct Point
{
    int x,y;
    Point(int x=0,int y=0):x(x),y(y){}
}point[N];
struct Line
{
    int x,y,z,pri;
    Line(int x=0,int y=0,int z=0,int pri=0):x(x),y(y),z(z),pri(pri){}
}line[N<<1];
bool cmpy(Point a,Point b)
{
    return a.y==b.y?a.x<b.x:a.y<b.y;
}
bool cmpx(Point a,Point b)
{
    return a.x==b.x?a.y<b.y:a.x<b.x;
}
bool cmpl(Line a,Line b)
{
    return a.x==b.x?a.pri<b.pri:a.x<b.x;
}
int A[N];
int main()
{
    // setIO("input");
    int i,j,n,cnt=0,ans=0;
    scanf("%d",&n);
    for(i=1;i<=n;++i)
    {
        scanf("%d%d",&point[i].x,&point[i].y);
        A[i]=point[i].y;
    }
    sort(A+1,A+1+n);
    for(i=1;i<=n;++i)
    {
        point[i].y=lower_bound(A+1,A+1+n,point[i].y)-A;
    }
    sort(point+1,point+1+n,cmpy);
    for(i=1;i<=n;i=j+1)
    {
        j=i;
        while(j<n&&point[j+1].y==point[i].y) ++j;
        line[++cnt]=Line(point[i].x,point[i].y,0,1);
        line[++cnt]=Line(point[j].x,point[j].y,0,3);
    }
    sort(point+1,point+1+n,cmpx);
    for(i=1;i<=n;i=j+1)
    {
        j=i;
        while(j<n&&point[j+1].x==point[i].x) ++j;
        line[++cnt]=Line(point[i].x,point[j].y,point[i].y,2);
    }
    sort(line+1,line+1+cnt,cmpl);
    for(i=1;i<=cnt;++i)
    {
        if(line[i].pri==2)
        {
            ans+=BIT::query(line[i].y)-BIT::query(line[i].z-1);
        }
        else
        {
            if(line[i].pri==1)  BIT::update(line[i].y,1);
            else BIT::update(line[i].y,-1);
        }
    }
    printf("%d\n",ans);
    return 0;
}