BOI2007 Mokia

分治套树状数组，板子题

#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a,ie=b;i<=ie;++i)
#define For(i,a,b) for(int i=a,ie=b;i<ie;++i)
using namespace std;
int read(){
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
    for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48);
    return x*f;
}
const int maxn=4e5+10;
typedef long long ll;

int ans[maxn],a[maxn];
struct oper{
    int x1,x2,y,f,id;//divide the y
    oper(){};
    oper(int x1,int x2,int y,int f,int id):
        x1(x1),x2(x2),y(y),f(f),id(id){};
}q[maxn];

bool cmp(oper a,oper b){
    if(a.y!=b.y)return a.y<b.y;
    return a.id<b.id;
}

struct TAS{
    int tr[maxn];
    int lb(int x){return x&-x;}
    void add(int x,int v){
        for(;x<maxn;x+=lb(x))tr[x]+=v;
    }
    int sum(int x){
        int ret=0;
        for(;x>0;x-=lb(x))ret+=tr[x];
        return ret;
    }
}T;

void solve(int l,int r){
    if(l==r)return;
    int mid=l+r>>1,j=0;
    solve(l,mid);solve(mid+1,r);
    vector<oper> s1,s2;
    s1.clear(),s2.clear();
    FOR(i,l,mid)if(!q[i].id)s1.push_back(q[i]);
    FOR(i,mid+1,r)if(q[i].id)s2.push_back(q[i]);
    sort(s1.begin(),s1.end(),cmp);
    sort(s2.begin(),s2.end(),cmp);
    For(i,0,s2.size()){
        for(int je=s1.size();j<je;++j)
            if(s1[j].y<=s2[i].y)
                T.add(s1[j].x1,s1[j].f);
            else break;
        ans[s2[i].id]+=s2[i].f*T.sum(s2[i].x2);
        ans[s2[i].id]-=s2[i].f*T.sum(s2[i].x1-1);
    }
    For(i,0,j)T.add(s1[i].x1,-s1[i].f);
}

int main(){
    int op=read(),n=read(),m=0,id_cnt=0;
    for(int x1,x2,y1,y2,v;(op=read())!=3;){
        if(op&1){
            x1=read();y1=read();v=read();
            q[++m]=oper(x1,x1,y1,v,0);a[m]=x1;
        }
        else{
            x1=read();y1=read();
            x2=read();y2=read();
            q[++m]=oper(x1,x2,y2,1,++id_cnt);a[m]=x2;
            q[++m]=oper(x1,x2,y1-1,-1,id_cnt);a[m]=x1-1;
        }
    }
    sort(a+1,a+m+1);
    int cnt=unique(a+1,a+m+1)-(a+1);
    for(int i=1;i<=m;++i){
        if(!q[i].id)
            q[i].x1=lower_bound(a+1,a+cnt+1,q[i].x1)-a;
        else{
            q[i].x1=lower_bound(a+1,a+cnt+1,q[i].x1-1)-a+1;
            q[i].x2=lower_bound(a+1,a+cnt+1,q[i].x2)-a;
        }
    }
    solve(1,m);
    for(int i=1;i<=id_cnt;++i)printf("%d
",ans[i]);
    return 0;
}