BZOJ 3110 [Zjoi2013]K大数查询 ——整体二分

Posted 2020-08-25

【题目分析】

    整体二分显而易见。

    自己YY了一下用树状数组区间修改，区间查询的操作。

    又因为一个字母调了一下午。

    貌似树状数组并不需要清空，可以用一个指针来维护，可以少一个log

    懒得写了。

【代码】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
  
using namespace std;
  
#define maxn 50005
#define inf 0x3f3f3f3f
#define ll long long 
  
ll n,m,cnt=0,tot=0;
  
struct Bit_Tree{
    ll a[maxn],b[maxn];
    void add(ll x,ll y,ll z)
    {
//      cout<<"Add "<<x<<" "<<y<<" "<<z<<endl;
        for (ll i=x;i<=n;i+=i&(-i)) b[i]+=z;
        for (ll i=y+1;i<=n;i+=i&(-i)) b[i]-=z;
        for (ll i=x;i<=n;i+=i&(-i)) a[i]+=(n-x)*z;
        for (ll i=y+1;i<=n;i+=i&(-i)) a[i]-=(n-y-1)*z;
    }
    void init()
    {
        memset(a,0,sizeof a);
        memset(b,0,sizeof b);
    }
    ll getsum(ll x)
    {
        ll ret=0,tmp=0;
        for (ll i=x;i;i-=i&(-i)) ret+=a[i],tmp+=b[i];
        return ret-(n-x-1)*tmp;
    }
    ll get(ll x,ll y)
    {
        return getsum(y)-getsum(x-1);
    }
}t;
  
struct data{
    ll opt,id;
    ll x,y,z;
}q[maxn<<1],q1[maxn<<1],q2[maxn<<1];
  
ll ans[maxn<<1],tag[maxn<<1];
  
void solve(ll ql,ll qr,ll l,ll r)
{
//  cout<<"solve"<<ql<<" "<<qr<<" "<<l<<" "<<r<<endl;
//  cout<<"In Que"<<endl;
    if (ql>qr) return;
    if (l==r)
    {
        for (ll i=ql;i<=qr;++i) ans[q[i].id]=l; 
        return ; 
    }
    ll mid=l+r>>1,p1=0,p2=0,cnt=0;
//    cout<<"Mid is "<<mid<<endl;
    for (ll i=ql;i<=qr;++i)
    {
//      cout<<q[i].opt<<" "<<q[i].x<<" "<<q[i].y<<" "<<q[i].z<<" "<<endl;
        if (q[i].opt==1)
        {
            if (q[i].z<=mid)
            {
                t.add(q[i].x,q[i].y,1);
                cnt+=q[i].y-q[i].x+1;
                q1[++p1]=q[i];
            }
            else q2[++p2]=q[i];
        }
        else
        {
            ll tmp=t.get(q[i].x,q[i].y);
//            cout<<"Tmp is "<<tmp<<endl;
            if (q[i].z<=tmp) q1[++p1]=q[i];
            else q[i].z-=tmp,q2[++p2]=q[i];
        }
    }
    for (ll i=1;i<=p1;++i)
    {
        if (q1[i].opt==1) t.add(q1[i].x,q1[i].y,-1);
        q[ql+i-1]=q1[i];
    }
    for (ll i=1;i<=p2;++i) q[ql+p1+i-1]=q2[i];
    solve(ql,ql+p1-1,l,mid);
    solve(ql+p1,qr,mid+1,r);
}
  
int main()
{
//  freopen("in.txt","r",stdin);
//  freopen("wa.txt","w",stdout);
    scanf("%lld",&n);
    n=maxn-1;
    scanf("%lld",&m);
    cnt=m;
    for (ll i=1;i<=m;++i)
    {
        scanf("%lld%lld%lld%lld",&q[i].opt,&q[i].x,&q[i].y,&q[i].z);
        q[i].id=i;
        if (q[i].opt==1) t.add(q[i].x,q[i].y,1);
        if (q[i].opt==2)
        {
            tag[i]=1;
            tot=t.get(q[i].x,q[i].y);
            q[i].z=tot-q[i].z+1;
        }
    }
    for (ll i=1;i<=m;++i)
        if (q[i].opt==1)
            t.add(q[i].x,q[i].y,-1);
    solve(1,cnt,-inf,inf);
    for (ll i=1;i<=m;++i)
        if (tag[i])
            printf("%lld\n",ans[i]);
}