洛谷$P2617\ Dynamic\ Rankings$ 整体二分

正解:整体二分

解题报告:

传送门$w$

阿查询带修区间第$k$小不显然整体二分板子呗,,,

就考虑先按时间戳排序(,,,其实并不需要读入的时候就按着时间戳排的鸭$QwQ$

每次二分出$mid$先把所有修改值小于等于$mid$的计入进去就成

查询的时候用值域线段树就成.

其实感觉和板子差不多,,,?而且修改还从区间变成单点了$QwQ$

$over$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define ri register int
#define rb register bool
#define rc register char
#define lowbit(x) (x&(-x))
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)

const int N=+,inf=1e9;
int n,m,a[N],nod_cnt,tr[N<<];
struct node{int l,r,K,id,as;}nod[N<<],tmp1[N<<],tmp2[N<<];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il bool cmp(node gd,node gs){return gd.id<gs.id;}
il bool rd(){rc ch=gc;while(ch!='Q' && ch!='C')ch=gc;return ch=='Q';}
il void modify(ri x,ri val){while(x<=n)tr[x]+=val,x+=lowbit(x);}
il int query(ri x){ri ret=;while(x)ret+=tr[x],x-=lowbit(x);return ret;}
void solv(ri l,ri r,ri dat_l,ri dat_r)
{
    if(l>r)return;
    if(dat_l==dat_r){rp(i,l,r)nod[i].as=dat_l;return;}
    ri mid=(dat_l+dat_r)>>,t1=,t2=;
    rp(i,l,r)
        if(nod[i].l>n)if(nod[i].K<=mid)tmp1[++t1]=nod[i],modify(nod[i].r,nod[i].l-n-);else tmp2[++t2]=nod[i];
        else
        {
            ri d=query(nod[i].r)-query(nod[i].l-);
            if(d>=nod[i].K)tmp1[++t1]=nod[i];else nod[i].K-=d,tmp2[++t2]=nod[i];
        }
    rp(i,l,r)if(nod[i].l>n && nod[i].K<=mid)modify(nod[i].r,-nod[i].l+n+);
    rp(i,,t1)nod[i+l-]=tmp1[i];rp(i,t1+,t1+t2)nod[i+l-]=tmp2[i-t1];
    solv(l,l+t1-,dat_l,mid);solv(l+t1,r,mid+,dat_r);
}

int main()
{
    //freopen("2617.in","r",stdin);freopen("2617.out","w",stdout);
    n=read();m=read();rp(i,,n)nod[++nod_cnt]=(node){n+,i,a[i]=read(),};
    rp(i,,m)
    {
        rb op=rd();ri pos=read();
        if(op)nod[++nod_cnt]=(node){pos,read(),read(),nod_cnt};
        else{nod[++nod_cnt]=(node){n+,pos,a[pos],};a[pos]=read();nod[++nod_cnt]=(node){n+,pos,a[pos],};}
    }
    solv(,nod_cnt,,inf);
    sort(nod+,nod+nod_cnt+,cmp);rp(i,,nod_cnt)if(nod[i].l<=n)printf("%d\n",nod[i].as);
    return ;
}

谁能想到我一个整体二分板子写了差不多两个小时呢,,,$QAQ$