poj 2104 划分树

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学了一下划分树，看着模板第一次写，挺好用的，就是不能动态更新。。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define MAXN 100010#define lson u<<1#define rson u<<1|1int sorted[MAXN],val[20][MAXN],Lnum[20][MAXN];int n,q;struct Node{int lef,rig,mid;}T[MAXN<<2];void Build(int u,int l,int r,int d){T[u].lef=l;T[u].rig=r;T[u].mid=(l+r)>>1;if(l==r)return;int mid=T[u].mid;int lsame=mid-l+1;for(int i=l;i<=r;i++){//记录进入左子树的个数，为了后面判断相等的时候要进入左子树还是右子树if(val[d][i]<sorted[mid])lsame--;}int p,q,same=0;p=T[u].lef;q=T[u].mid+1;for(int i=l;i<=r;i++){if(i==l)Lnum[d][i]=0;else Lnum[d][i]=Lnum[d][i-1];if(val[d][i]<sorted[mid]){val[d+1][p++]=val[d][i];Lnum[d][i]++;}else if(val[d][i]>sorted[mid]){val[d+1][q++]=val[d][i];}else {if(same<lsame){same++;val[d+1][p++]=val[d][i];Lnum[d][i]++;}else val[d+1][q++]=val[d][i];}}Build(lson,l,mid,d+1);Build(rson,mid+1,r,d+1);}int Query(int u,int l,int r,int d,int k){if(l==r)return val[d][l];int n1,n2;//n2记录[l,r]中进入左子树的个数if(l==T[u].lef){n1=0;n2=Lnum[d][r];}else {n1=Lnum[d][l-1];n2=Lnum[d][r]-n1;}if(n2>=k){int newl=T[u].lef+n1;int newr=T[u].lef+n1+n2-1;return Query(lson,newl,newr,d+1,k);}else {int newl=T[u].mid+1+(l-T[u].lef-n1);int newr=newl+(r-l+1-n2)-1;return Query(rson,newl,newr,d+1,k-n2);}}int main(){while(scanf("%d%d",&n,&q)==2){for(int i=1;i<=n;i++){scanf("%d",&val[0][i]);sorted[i]=val[0][i];}sort(sorted+1,sorted+1+n);Build(1,1,n,0);int l,r,k;while(q--){scanf("%d%d%d",&l,&r,&k);printf("%d\n",Query(1,l,r,0,k));}}return 0;}