BZOJ 1176: [Balkan2007]Mokia/2683: 簡單題

CDQ入門題 還雙倍經驗QAQ
我怎麼好像對CDQ生疏了。。。

拆成x1-1 y1~y2和 x2 y1~y2
然後分治一下 再加個樹狀數組來算y就好了。。。

感覺我（的代碼）又要被人說又短又快了QAQ

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof a)
using namespace std;
typedef long long LL;
const int N=2e6+2,M=4e5+2;
char B[1<<14],*S=B,*T=B;
#define gc (S==T&&(T=(S=B)+fread(B,1,1<<14,stdin),S==T)?-1:*S++)
inline int read(){
    int x=0,f=1; char ch=gc;
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1; ch=gc;}
    while(ch>='0' && ch<='9'){x=(x<<1)+(x<<3)+ch-'0'; ch=gc;}
    return x*f;
}
struct ask{
    int x,y1,y2,p;
}q[M],t[M]; int ql;
bool Cmp(ask x1,ask x2){
    return x1.x<x2.x || (x1.x==x2.x && x1.p==0 && x2.p!=0);
}
int n,ans[M>>1],s[N];
void add(int x,int u){for(;x<=n;x+=x&-x)s[x]+=u;}
int sum(int x){int u=0; for(;x;x-=x&-x)u+=s[x]; return u;}
void cdq(int l,int r){
    if(l==r)return;
    int mid=l+r>>1,i;
    cdq(l,mid),cdq(mid+1,r);
    int l1=l,l2=mid+1;
    for(;l2<=r;++l2){
        for(;q[l1].x<=q[l2].x && l1<=mid;++l1) if(!q[l1].p)
            add(q[l1].y1,q[l1].y2);
        if(q[l2].p!=0){
            if(q[l2].p<0) ans[-q[l2].p]-= sum(q[l2].y2)-sum(q[l2].y1-1);
            else ans[q[l2].p]+= sum(q[l2].y2)-sum(q[l2].y1-1);
        }
    }
    for(i=l;i<l1;++i) if(!q[i].p)
        add(q[i].y1,-q[i].y2);
    int p=l; l1=l,l2=mid+1;
    for(i=l;i<=r;++i){
        if( l1<=mid && (l2>r || Cmp(q[l1],q[l2]) ) )t[p++]=q[l1++];
        else t[p++]=q[l2++];
    }
    for(i=l;i<=r;++i)q[i]=t[i];
}
int main(){
    int u=read(),o=0; n=read();
    int r=read();
    while(r!=3){
        if(r==1){
            int x=read(),y=read(),c=read();
            q[++ql]=(ask){x,y,c,0};
        }
        else{
            int x1=read(),y1=read(),x2=read(),y2=read();
            ++o; ans[o]=(y2-y1+1)*(x2-x1+1)*u;
            q[++ql]=(ask){x1-1,y1,y2,-o};
            q[++ql]=(ask){x2,y1,y2,o};
        }
        r=read();
    }
    cdq(1,ql);
    for(int i=1;i<=o;++i) printf("%d\n",ans[i]);
    return 0;
}