BZOJ 3110 [Zjoi2013]K大数查询 ——树套树

【题目分析】

    外层区间线段树，内层是动态开点的权值线段树。

    SY神犇说树套树注重的是内外层的数据结构的选择问题，果然很重要啊。

    动态开点的实现方法很好。

【代码】

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>

#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <vector>
#include <iostream>
#include <queue>

using namespace std;

#define mlog 16
#define inf (0x3f3f3f3f)

unsigned Getunsigned()
{
    unsigned x=0,f=1; char ch=getchar();
    while (ch<'0'||ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0'&&ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;

}

#define maxn 50005
#define F(i,j,k) for (unsigned i=j;i<=k;++i)
#define maxm maxn<<8

unsigned rt[maxn<<2],n,m,sum[maxm],le[maxm],ri[maxm],lazy[maxm];
unsigned c,L,R,cnt,opt;

void push(unsigned &o,unsigned l,unsigned r)
{
    if (!o) o=++cnt;
    if (L<=l&&r<=R)
    {
        lazy[o]++;
        sum[o]+=(r-l+1);
        return;
    }
    unsigned mid=l+r>>1;
    if (L<=mid) push(le[o],l,mid);
    if (R>mid) push(ri[o],mid+1,r);
    sum[o]=sum[le[o]]+sum[ri[o]]+lazy[o]*(r-l+1);
}

void modi(unsigned o,unsigned l,unsigned r)
{
    push(rt[o],1,n); if (l==r) return ;
    unsigned mid=l+r>>1;
    if (c<=mid) modi(o<<1,l,mid);
    else modi(o<<1|1,mid+1,r);
}

unsigned calc(unsigned o,unsigned l,unsigned r)
{
    if (!o) return 0;
    if (L<=l&&r<=R) return sum[o];
    unsigned mid=l+r>>1,tmp=0;
    if (L<=mid) tmp+=calc(le[o],l,mid);
    if (R>mid) tmp+=calc(ri[o],mid+1,r);
    return tmp+lazy[o]*(min(R,r)-max(L,l)+1);
}

unsigned ask(unsigned o,unsigned l,unsigned r)
{
    if (l==r) return l;
    unsigned mid=l+r>>1,tmp=calc(rt[o<<1],1,n);
    if (c<=tmp) return ask(o<<1,l,mid);
    c-=tmp; return ask(o<<1|1,mid+1,r);
}

int main()
{
    n=Getunsigned();
    m=Getunsigned();
    F(i,1,m)
    {
        opt=Getunsigned(); L=Getunsigned();
        R=Getunsigned(); c=Getunsigned();
        if (opt==1) c=n-c+1,modi(1,1,n);
        else printf("%u
",n-ask(1,1,n)+1);
    }
    return 0;
}