Kth number (HDU_2665) 划分树

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7845    Accepted Submission(s): 2440

Problem Description

Give you a sequence and ask you the kth big number of a inteval.

 

Input

The first line is the number of the test cases. 
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere. 
The second line contains n integers, describe the sequence. 
Each of following m lines contains three integers s, t, k. 
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

 

Output

For each test case, output m lines. Each line contains the kth big number.

 

Sample Input

1 10 1 1 4 2 3 5 6 7 8 9 0 1 3 2 

 

Sample Output

2
题目大意：求kth值.
解题思路：参考POJ_2104,99%相同。
代码如下：
#include"cstdio"#include"iostream"#include"algorithm"#define MAXN 100000 + 10using namespace std;struct seq{int order;int value;}a[MAXN];struct node{//划分树 int val[MAXN];//存值 int left[MAXN];//划入左子树的个数 int deep;//深度 }tree[25];bool cmp(seq a,seq b){return a.value<b.value;} //构建划分树 void Build(int d,int l,int r){if(l==r) return ;int mid=(l+r)>>1;int p=0,t=0;for(int i=l;i<=r;i++){if(tree[d].val[i]<=mid){//划入左子树 tree[d+1].val[l+p]=tree[d].val[i];p++;tree[d].left[i]=p; }else{//划入右子树 tree[d+1].val[mid+1+t]=tree[d].val[i];t++;tree[d].left[i]=p; //是p不是t }}Build(d+1,l,mid);Build(d+1,mid+1,r);}//在大区间(l,r)中查找小区间(ll,rr)的第k大值 int Query(int d,int l,int r,int ll,int rr,int k){ if(ll==rr) return (tree[d].val[ll]);int ls=0,rs=0;//ls,rs用于确定小区间if(ll>l)ls=tree[d].left[ll-1];else ls=0;int mid=(l+r)>>1;rs=tree[d].left[rr];if(rs-ls>=k) return Query(d+1,l,mid,l+ls,l+rs-1,k);else return Query(d+1,mid+1,r,(mid+1)+(ll-l-ls),(mid+1)+rr-l-rs,k-(rs-ls)); if(ll==rr) return tree[d].val[ll];}int main(){int n,m;int i;int T;scanf("%d",&T); while(T--){scanf("%d%d",&n,&m); //存值并进行离散化 for(i=1;i<=n;i++){scanf("%d",&a[i].value);a[i].order=i;}sort(a+1,a+1+n,cmp);for(i=1;i<=n;i++){tree[0].val[a[i].order]=i;}Build(0,1,n);while(m--){int ll,rr,k;scanf("%d%d%d",&ll,&rr,&k);printf("%d\n",a[Query(0,1,n,ll,rr,k)].value);}}return 0;}