HDU 4006 The kth great number（线段树【亚洲区网络赛题目】）

The kth great number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 9217 Accepted Submission(s): 3652

Problem Description
Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.

Input
There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an ” I” followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a “Q”, then you need to output the kth great number.

Output
The output consists of one integer representing the largest number of islands that all lie on one line.

Sample Input
8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q

本题的意思是：让你找出已经写入的数字的第k大数字是哪个。就是正过来的第（sum-k+1）个数字是哪个。

下面是AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 1000005
using namespace std;

struct node
{
int left,right,val;
} c[maxn*3];

void build_tree(int root,int l,int r)//建树
{
    c[root].left=l;
    c[root].right=r;
    c[root].val=0;
if(c[root].left==c[root].right)
    {
return ;
    }
int mid=(c[root].left+c[root].right)/2;
    build_tree(root*2,l,mid);
    build_tree(root*2+1,mid+1,r);
    c[root].val=c[root*2].val+c[root*2+1].val;
}

void update_tree(int root,int pos,int l,int r)//单点更新和区间更新
{
if(l==r)
    {
        c[root].val++;
return;
    }
int mid=(c[root].left+c[root].right)/2;
if(pos>mid)
    {
        update_tree(root*2+1,pos,mid+1,r);
    }
else
    {
        update_tree(root*2,pos,l,mid);
    }
    c[root].val=c[root*2].val+c[root*2+1].val;//区间更新
}

int search_tree(int root,int pos,int l,int r)//点的查找
{
int re;
if(l==r)//把找到的数返回
    {
return l;
    }
else
    {
int mid=(c[root].left+c[root].right)/2;
if(pos<=c[root*2].val)
        {
            re=search_tree(root*2,pos,l,mid);
        }
else
        {
            re=search_tree(root*2+1,pos-c[root*2].val,mid+1,r);
        }
    }
return re;
}

int main()
{
int n,k;
while(~scanf("%d%d",&n,&k))
    {
int sum=0;//记录已经出现的数字的个数
        build_tree(1,1,maxn);
        char str[3];
int num;
for(int i=0; i<n; i++)
        {
            scanf("%s",str);
if(str[0]=='I')
            {
                sum++;
                scanf("%d",&num);
                update_tree(1,num,1,n);
            }
else if(str[0]=='Q')
            {
int re=search_tree(1,sum-k+1,1,maxn);
printf("%d\n",re);
            }
        }
    }
return 0;
}