BZOJ1935 园丁的烦恼

一个二维偏序的问题，学过了三维偏序cdq分治之后觉得这个题非常的水。只需按一维排序之后再用树状数组操作即可。——by VANE

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+;
struct qry
{
    int x,y,id;
}q[N<<];
struct tree
{
    int x,y;
}t[N];
int s[N*],n,m,tmp[N*],num,vis_tim,ans[N*];
bool cmp(qry a,qry b)
{
    return a.x<b.x;
}
bool cmpp(tree a,tree b)
{
    return a.x<b.x;
}
void add(int x)
{
    for(;x<=num;x+=x&-x)
        s[x]++;
}
int query(int x)
{
    int res=;
    for(;x;x-=x&-x)
        res+=s[x];
    return res;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=;i<=n;++i)
        scanf("%d%d",&t[i].x,&t[i].y);
    for(int i=;i<=m;++i)
    {
        int x,y,xx,yy;scanf("%d%d%d%d",&x,&y,&xx,&yy);
        int pos=(i-)*;
        q[pos+].x=xx;q[pos+].y=yy;q[pos+].id=pos+;
        q[pos+].x=x-;q[pos+].y=y-;q[pos+].id=pos+;
        q[pos+].x=xx;q[pos+].y=y-;q[pos+].id=pos+;
        q[pos+].x=x-;q[pos+].y=yy;q[pos+].id=pos+;
    }
    for(int i=;i<=n;++i)
    tmp[i]=t[i].y;
    for(int i=;i<=m*;++i)
    tmp[i+n]=q[i].y;
    sort(tmp+,tmp++n+m*);
    num=unique(tmp+,tmp++n+*m)-tmp-;
    for(int i=;i<=n;++i) t[i].y=lower_bound(tmp+,tmp++num,t[i].y)-tmp;
    for(int i=;i<=m*;++i) q[i].y=lower_bound(tmp+,tmp++num,q[i].y)-tmp;
    sort(q+,q++*m,cmp);
    sort(t+,t++n,cmpp);
    int i=,j=;
    while(j<=*m)
    {
        int h=q[j].x;
        while(t[i].x<=h&&i<=n) add(t[i].y),++i;
        while(j<=*m&&q[j].x==h)
        {
            ans[q[j].id]+=query(q[j].y);
            ++j;
        }
    }
    for(int i=;i<=*m;i+=)
        printf("%d\n",ans[i]+ans[i+]-ans[i+]-ans[i+]);
}

二维偏序用cdq我是不是有病，我就是要写cdq 15000ms卡过去了

By：大奕哥

 #include<bits/stdc++.h>
 using namespace std;
 struct node
 {
     int x,y,w,id,f;
     bool operator <(const node &b)const
     {
         return x==b.x?y<b.y:x<b.x;
     }
 }q[];
 int cnt,maxr,tim;
 int t[],v[],ans[],pos[];
 inline int lowbit(int x){return x&(-x);}
 void add(int x,int w)
 {
     for(;x<=maxr;x+=lowbit(x))
     {
         if(v[x]!=tim)
         {
             v[x]=tim;t[x]=w;
         }
         else t[x]+=w;
     }
 }
 int query(int x)
 {
     int an=;
     for(;x;x-=lowbit(x))
     if(v[x]==tim)
     an+=t[x];
     return an;
 }
 void cdq(int l,int r)
 {
     if(l==r)return;
     int mid=(l+r)>>;
     cdq(l,mid);cdq(mid+,r);
     sort(q+l,q+mid+);sort(q+mid+,q+r+);
     tim++;
     int i=l,j=mid+;
     while(j<=r)
     {
         while(q[i].f==&&i<=mid)++i;
         while(q[j].f==&&j<=r)++j;
         if(q[i].x<=q[j].x&&i<=mid)add(q[i].y,q[i].w),++i;
         else if(j<=r)ans[q[j].id]+=query(q[j].y),++j;
     }
 }
 int main()
 {
     int n,m;
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;++i)
     {
         int x,y;
         scanf("%d%d",&x,&y);x++;y++;
         q[++cnt].f=;q[cnt].x=x;q[cnt].y=y;q[cnt].w=;q[cnt].id=cnt;maxr=max(maxr,y);
     }
     for(int i=;i<=m;++i)
     {
         int x1,x2,y1,y2;
         scanf("%d%d%d%d",&x1,&y1,&x2,&y2);x1++;x2++;y1++;y2++;
         pos[i]=cnt;maxr=max(maxr,max(y1,y2));
         q[++cnt].f=;q[cnt].x=x2;q[cnt].y=y2;q[cnt].id=cnt;
         q[++cnt].f=;q[cnt].x=x1-;q[cnt].y=y1-;q[cnt].id=cnt;
         q[++cnt].f=;q[cnt].x=x1-;q[cnt].y=y2;q[cnt].id=cnt;
         q[++cnt].f=;q[cnt].x=x2;q[cnt].y=y1-;q[cnt].id=cnt;
     }
     cdq(,cnt);
     for(int i=;i<=m;++i)
     printf("%d\n",ans[pos[i]+]+ans[pos[i]+]-ans[pos[i]+]-ans[pos[i]+]);
     return ;
 }