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 + 10
using 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;
}