POJ2104 K-th Number [整体二分]

//It is made by HolseLee on 5th Oct 2018
//POJ2104
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;

const int N=2e5+7;
int n,m,ans[N],cnt,c[N];
struct Node {
    int x,y,k,pos,type;
    Node() {}
    Node(const int _x,const int _y,const int _k,const int _p,const int _t):
        x(_x), y(_y), k(_k), pos(_p), type(_t) {}
}q[N],q1[N],q2[N];

inline int read()
{
    char ch=getchar(); int num=0; bool flag=false;
    while( ch<‘0‘ || ch>‘9‘ ) {
        if( ch==‘-‘ ) flag=true; ch=getchar();
    }    
    while( ch>=‘0‘ && ch<=‘9‘ ) {
        num=num*10+ch-‘0‘; ch=getchar();
    }
    return flag ? -num : num;
}

inline int lowbit(int x)
{
    return x&(-x);
}

inline void add(int pos,int x)
{
    for(; pos<=n; pos+=lowbit(pos)) c[pos]+=x;
}

inline int quary(int pos)
{
    int ret=0;
    for(; pos>0; pos-=lowbit(pos)) ret+=c[pos];
    return ret;
}

void solve(int l,int r,int L,int R)
{
    if( l>r || L>R ) return;
    if( l==r ) {
        for(int i=L; i<=R; ++i) 
        if( q[i].type ) ans[q[i].pos]=l;
        return ;
    }
    int mid=(l+r)>>1, cnt1=0, cnt2=0;
    for(int i=L; i<=R; ++i) 
    if( q[i].type ) {
        int tmp=quary(q[i].y)-quary(q[i].x-1);
        if( tmp>=q[i].k ) q1[++cnt1]=q[i];
        else q[i].k-=tmp, q2[++cnt2]=q[i];
    } else {
        if( q[i].x<=mid ) q1[++cnt1]=q[i], add(q[i].pos,1);
        else q2[++cnt2]=q[i];
    }
    for(int i=1; i<=cnt1; ++i) 
    if(!q1[i].type) add(q1[i].pos,-1);
    for(int i=1; i<=cnt1; ++i) q[L+i-1]=q1[i];
    for(int i=1; i<=cnt2; ++i) q[L+cnt1+i-1]=q2[i];
    solve(l,mid,L,L+cnt1-1), solve(mid+1,r,L+cnt1,R);
}

int main()
{
    n=read(), m=read();
    int x,y,z;
    for(int i=1; i<=n; ++i) 
        x=read(), q[++cnt]=Node(x,1,0,i,0);
    for(int i=1; i<=m; ++i) {
        x=read(), y=read(), z=read();
        q[++cnt]=Node(x,y,z,i,1);
    }
    solve(-inf,inf,1,cnt);
    for(int i=1; i<=m; ++i) 
        printf("%d\n",ans[i]);
    return 0;
}