HDOJ 题目2852 KiKi's K-Number（线段树求大于a的第k值）

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本文概述了AI音视频处理领域的关键技术，包括视频分割、语义识别、自动驾驶、AR、SLAM等，并探讨了其在实际应用中的作用。

KiKi's K-Number

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3258 Accepted Submission(s): 1459

Problem Description

For the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the container is to support the three operations.

Push: Push a given element e to container

Pop: Pop element of a given e from container

Query: Given two elements a and k, query the kth larger number which greater than a in container;

Although Kiki is very intelligent, she can not think of how to do it, can you help her to solve this problem?

Input

Input some groups of test data ,each test data the first number is an integer m (1 <= m <100000), means that the number of operation to do. The next m lines, each line will be an integer p at the beginning, p which has three values:
If p is 0, then there will be an integer e (0 <e <100000), means press element e into Container.

If p is 1, then there will be an integer e (0 <e <100000), indicated that delete the element e from the container

If p is 2, then there will be two integers a and k (0 <a <100000, 0 <k <10000),means the inquiries, the element is greater than a, and the k-th larger number.

Output

For each deletion, if you want to delete the element which does not exist, the output "No Elment!". For each query, output the suitable answers in line .if the number does not exist, the output "Not Find!".

Sample Input

Sample Output

  
   No Elment!
6
Not Find!
2
2
4
Not Find!

Source

2009 Multi-University Training Contest 4 - Host by HDU

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题目大意：三种操作0表示加入一个数，1表示删除一个数，2表示查询比a大的第k个数

15384508

2015-11-04 21:12:23

Accepted

2852

1014MS

3120K

2695 B

C++

XY_

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define n 100001
#define LL long long
using namespace std;
int node[n<<2];
int vis[n];
void pushup(int tr)
{
    node[tr]=node[tr<<1]+node[tr<<1|1];
}
void build(int l,int r,int tr)
{
    node[tr]=0;
    if(l==r)
    {
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,tr<<1);
    build(mid+1,r,tr<<1|1);
}
void update(int pos,int val,int l,int r,int tr)
{
    if(l==r)
    {
        node[tr]+=val;
        return;
    }
    int mid=(l+r)>>1;
    if(pos<=mid)
    {
        update(pos,val,l,mid,tr<<1);
    }
    else
        update(pos,val,mid+1,r,tr<<1|1);
    pushup(tr);
}
int query(int L,int R,int l,int r,int tr)
{
    if(L==l&&r==R)
    {
        return node[tr];
    }
    int mid=(l+r)>>1;
    if(R<=mid)
        return query(L,R,l,mid,tr<<1);
    else
        if(L>mid)
            return query(L,R,mid+1,r,tr<<1|1);
        else
        {
            int a=query(L,mid,l,mid,tr<<1);
            int b=query(mid+1,R,mid+1,r,tr<<1|1);
            return a+b;
        }
}
int find(int num,int l,int r,int tr)
{
    if(node[tr]<num)
        return -1;
    if(l==r)
        return l;
    int mid=(l+r)>>1;
    if(node[tr<<1]>=num)
        find(num,l,mid,tr<<1);
    else
        find(num-node[tr<<1],mid+1,r,tr<<1|1);
}
int main()
{
    int q;
    while(scanf("%d",&q)!=EOF)
    {
        build(1,n,1);
        memset(vis,0,sizeof(vis));
        while(q--)
        {
            int op;
            scanf("%d",&op);
            if(op==0)
            {
                int x;
                scanf("%d",&x);
                update(x,1,1,n,1);
                vis[x]++;
            }
            else
            {
                if(op==1)
                {
                    int x;
                    scanf("%d",&x);
                    if(vis[x]==0)
                        printf("No Elment!\n");
                    else
                    {
                        update(x,-1,1,n,1);
                        vis[x]--;
                    }
                }
                else
                {
                    int a,b;
                    scanf("%d%d",&a,&b);
                    int num=query(1,a,1,n,1);
                   // printf("+++++%d\n",num);
                    num+=b;
                  //  printf("++++++%d\n",num);
                    int c=find(num,1,n,1);
                    if(c==-1)
                        printf("Not Find!\n");
                    else
                        printf("%d\n",c);
                }
            }
        }
    }
}