HDU 2665.Kth number 区间第K小

Posted 2020-09-28 GeekZRF

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11394    Accepted Submission(s): 3465

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）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2665

题意：求区间第K小（区间升序第k个数）

思路：归并树。二分答案x。判断不超过x的数在区间内多少个。假设x是区间第k小，那么在区间内不超过x的数不少于k个。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
using namespace std;
#define PI acos(-1.0)
typedef long long ll;
typedef pair<int,int> P;
const int maxn=1e5+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e13+7;
inline int get_int()
{
    int num=0;
    char ch;
    while((ch=getchar())!=‘ ‘&&ch!=‘\n‘)
        num=num*10+(ch-‘0‘);
    return num;
}
/****************************/
struct edge
{
    int from,to;
    int cost;
};
edge es[maxm];
struct node
{
    int num;
    int k;
};
node sign[maxn];
int a[maxn];
int cmp(node x,node y)
{
    return x.num<y.num;
}
vector<node>tree[maxn<<2];
void build(int l,int r,int pos)
{
    if(l==r) return;
    int mid=(l+r)/2;
    for(int i=0; i<tree[pos].size(); i++)
    {
        int k=tree[pos][i].k;
        if(l<=k&&k<=mid) tree[pos<<1].push_back(tree[pos][i]);
        else tree[pos<<1|1].push_back(tree[pos][i]);
    }
    build(l,mid,pos<<1);
    build(mid+1,r,pos<<1|1);
}
int query(int L,int R,int w,int l,int r,int pos)
{
    if(L<=l&&r<=R)
    {
        int s=0,e=tree[pos].size()-1;
        int cou=-1;
        while(s<=e)
        {
            int md=(s+e)/2;
            if(tree[pos][md].num<=w) s=md+1,cou=md;
            else e=md-1;
        }
        return cou+1;
    }
    int mid=(l+r)/2;
    int ans=0;
    if(L<=mid) ans+=query(L,R,w,l,mid,pos<<1);
    if(R>mid) ans+=query(L,R,w,mid+1,r,pos<<1|1);
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,q;
        scanf("%d%d",&n,&q);
        for(int i=0; i<=4*n; i++) tree[i].clear();
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
            sign[i].num=a[i],sign[i].k=i+1;
        }
        sort(sign,sign+n,cmp);
        for(int i=0; i<n; i++) tree[1].push_back(sign[i]);
        build(1,n,1);
        while(q--)
        {
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            int L=0,R=n-1;
            int ans=n-1;
            while(L<=R)
            {
                int mid=(L+R)/2;
                int w=sign[mid].num;
                if((query(l,r,w,1,n,1))>=k) R=mid-1,ans=mid;
                else L=mid+1;
            }
            printf("%d\n",sign[ans].num);
        }
    }
    return 0;
}

区间第K小