HDU 2665 Kth number （主席树静态查询区间第k小）

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18910    Accepted Submission(s): 5742

 

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

 

 

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

 

 

Recommend

zty

 

主席树的模板题

/*静态查询区间第k小*/
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll ;
const int oo=0x7f7f7f7f ;
const int maxn=1e5+7;
const int mod=1e9+7;
int t,n,m,cnt,root[maxn],a[maxn],x,y,k;
//cnt和root：主席树的总点数和每一个根
struct node{
    int l,r,sum;
}T[maxn*25];
vector v;
int getid(int x){ //离散化
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
///单点修改
void update(int l,int r,int &x,int y,int pos){
    T[++cnt]=T[y],T[cnt].sum++,x=cnt;
    if(l==r) return;
    int mid=(l+r)/2;
    if(mid>=pos) update(l,mid,T[x].l,T[y].l,pos);
    else update(mid+1,r,T[x].r,T[y].r,pos);
}
///查询区间（x，y）第k小
int query(int l,int r,int x,int y,int k){
    if(l==r) return l;
    int mid=(l+r)/2;
    int sum=T[T[y].l].sum-T[T[x].l].sum;
    if(sum>=k) return query(l,mid,T[x].l,T[y].l,k);
    else return query(mid+1,r,T[x].r,T[y].r,k-sum);
}
int main(void){
    scanf("%d",&t);
    while(t--){
        v.clear();
        cnt=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]),v.push_back(a[i]);
        sort(v.begin(),v.end());
        v.erase(unique(v.begin(),v.end()),v.end());
 
        for(int i=1;i<=n;i++)
        	 update(1,n,root[i],root[i-1],getid(a[i]));
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&x,&y,&k);
            printf("%d\n",v[query(1,n,root[x-1],root[y],k)-1]);
        }
    }
	return 0 ;
}