Mokia (three-dimensional partial order) P4390

Mentioned cdq, would have to mention this damn road, suffer a thousand knives topic.

Minimalist questions surface:

Given a two-dimensional plane, at time ti will put a point (xi, yi), queries in a block which point the number of time tj

See questions is a two-dimensional tree line fucks ah, so the water? ? ?

Discouraging data range warning

Is a one-dimensional array of almost a bit too much ... 1e6 * 1e6's? Several GB? ? ?

。。。

So, the great divide and conquer CDQ played.

In fact, the translation may face problems like this:

Insertion point in time, to ask less than (x, y) and the time point the current point is less than the number

Is not that CDQ thing? Even more than the bare template title. . .

But you may want to look at the difference (two-dimensional differential) because the statistics is the large rectangle consisting of 00 points, so to cut two rectangles, plus a small rectangle, so to count partial order four points

To sum up, it is cdq.

A first time dimension, a second dimension x, y of the third dimension

Be sure to do offline

So began a pleasant CDQ

#include<bits/stdc++.h>
using namespace std;
const int maxn=505;

struct node
{
    int time,x,y,val,id;
}e[maxn];
int m,cnt,t[maxn<<1],a[maxn],ans[maxn];
inline int lowbit(int x)
{
    return x & - x ;
}
void add(int x,int y)
{
    for(;x<=m;x+=lowbit(x))
    {
        t[x]+=y;
    }
}
int ask(int x)
{
    int res=0;
    for(;x;x-=lowbit(x))
    {
        res+=t[x];
    }
    return res;
}
bool cmp2(node a,node b)
{
    if(a.x!=b.x)return a.x<b.x;
    if(a.y!=b.y)return a.y<b.y;
    //else        return a.time<b.time;
}
bool cmp(node a,node b)
{
    return a.time<b.time;
}
void cdq(int l,int r)
{
    if(l==r)return;
    int mid=l+r>>1;
    cdq(l,mid);
    cdq(mid+1,r);
    sort(e+l,e+1+r,cmp2);
    for(int i=l;i<=r;i++)
    {
        if(e[i].x<=mid&&e[i].id==0)
        add(e[i].y,e[i].val);
        else e[i].val+=ask(e[i].y);
    }
    for(int i=l;i<=r;i++)
    {
        if(e[i].x<=mid&&e[i].id==0)
        add(e[i].y,-e[i].val);
    }
}
int read()
{
    int f=1,x=0;char s=getchar();
    while(s>'9'||s<'0'){if(s=='-')f=-1;s=getchar();}
    while(s<='9'&&s>='0'){x=x*10+s-'0';s=getchar();}
    return x*f;
}
int main()
{
    read();
    m=read();
    int flag=read();
    while(flag!=3)
    {
        if(flag==1)
        {
            int x=read()+1,y=read()+1,val=read();
            e[++cnt]=(node){cnt,x,y,val,0};
        }
        else
        {
            intRead = X1 (), YL = Read (), X2 = Read () + . 1 , Y2 = Read () + . 1 ; 
            E [ ++ CNT] = (Node) {CNT, X1, YL, 0 , . 1 }; // data structure of 
            E [ ++ CNT] = (Node) {CNT, X2, Y2, 0 , . 1 }; 
            E [ ++ CNT] = (Node) {CNT, X2, YL, 0 , . 1 }; 
            E [ ++ CNT] = (Node) {CNT, X1, Y2, 0 , . 1 }; 
        } 
        In Flag = Read (); 
    } 
    cdq ( . 1 , CNT); and hard cdq on the line 
    Sort (E + . 1 , E + + cnt 1,cmp);
    for(int i=1;i<=cnt;++i)
    {
        if(e[i].id==1)
        {
            printf("%d\n",e[i].val+e[i+1].val-e[i+2].val-e[i+3].val);
            i+=3;
        }
    }
    return 0;
}

(Finish)