整体二分--BZOJ1901: Zju2112 Dynamic Rankings

n<=10000个数有m<=10000个操作，1、询问一个区间的第k小的数；2、单点修改。

带修主席树。

整体二分。

整体二分的必要条件：

 #include<string.h>
 #include<stdlib.h>
 #include<stdio.h>
 //#include<assert.h>
 #include<algorithm>
 //#include<iostream>
 using namespace std;

 int n,m;
 #define maxn 30011
 const int inf=0x3f3f3f3f;
 struct Ope
 {
     int x,y,z,id,type;
     //type=1: 询问区间[x,y]第z大，询问编号id
     //type=0：把位置x的数修改为y，增加：z=1，删除：z=-1
 }q[maxn],al[maxn],ar[maxn];int len;
 int ans[maxn];

 struct BIT
 {
     int a[maxn];
     void add(int x,int v) {for (;x<=n;x+=x&-x) a[x]+=v;}
     int query(int x) {int ans=; for (;x;x-=x&-x) ans+=a[x]; return ans;}
 }t;

 void solve(int L,int R,int ql,int qr)
 {
     if (L>R || ql>qr) return;
     if (L==R)
     {
         for (int i=ql;i<=qr;i++) if (q[i].type) ans[q[i].id]=L;
         return;
     }
     int mid=(L+R)>>,lql=,lqr=;
     for (int i=ql;i<=qr;i++)
     {
         if (q[i].type)
         {
             int tmp=t.query(q[i].y)-t.query(q[i].x-);
             if (tmp>=q[i].z) al[++lql]=q[i];
             else
             {
                 q[i].z-=tmp;
                 ar[++lqr]=q[i];
             }
         }
         else
         {
             if (q[i].y<=mid)
             {
                 t.add(q[i].x,q[i].z);
                 al[++lql]=q[i];
             }
             else ar[++lqr]=q[i];
         }
     }
     for (int i=;i<=lql;i++) if (al[i].type==) t.add(al[i].x,-al[i].z);
     for (int i=,j=ql;i<=lql;i++,j++) q[j]=al[i];
     for (int i=,j=ql+lql;i<=lqr;i++,j++) q[j]=ar[i];
     solve(L,mid,ql,ql+lql-);
     solve(mid+,R,ql+lql,qr);
 }

 int a[maxn];
 int main()
 {
     scanf("%d%d",&n,&m);
     for (int i=;i<=n;i++) scanf("%d",&q[i].y),a[i]=q[i].y,q[i].x=i,q[i].z=,q[i].type=;
     char s[];
     int cntq=;
     for (int i=,j=n+;i<=m;i++,j++)
     {
         scanf("%s",s);
         if (s[]=='Q') scanf("%d%d%d",&q[j].x,&q[j].y,&q[j].z),q[j].id=++cntq,q[j].type=;
         else scanf("%d%d",&q[j].x,&q[j].y),q[j].z=,q[j].type=,
         q[j+]=(Ope){q[j].x,a[q[j].x],-,,},a[q[j].x]=q[j].y,j++;
         if (i==m) len=j;
     }
     solve(,inf,,len);
     for (int i=;i<=cntq;i++) printf("%d\n",ans[i]);
     return ;
 }