hdu 2665 Kth number（主席树）

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15329    Accepted Submission(s): 4600

 

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.

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Sample Input

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

Sample Output

2

题意：求区间第K大的数

思路：主席树的模版题（大家可以看这个博主的视频，我感觉讲的不错）

代码：

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int cnt,root[N],a[N];
int n,m;
struct node
{
    int l,r,sum;
}T[N*40];
vector<int>v;
int getid(int x)
{
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
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(pos<=mid)update(l,mid,T[x].l,T[y].l,pos);
    else update(mid+1,r,T[x].r,T[y].r,pos);
}
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()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        cnt=0;
        memset(root,0,sizeof(root));
        v.clear();
        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++)
        {
            int x,y,k;
            scanf("%d%d%d",&x,&y,&k);
            printf("%d\n",v[query(1,n,root[x-1],root[y],k)-1]);
        }
    }
    return 0;
}