HDU 4006 The kth great number

The kth great number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

Total Submission(s): 10799    Accepted Submission(s): 4307


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

 

Sample Output

1
2
3

HintXiao Ming won't ask Xiao Bao the kth great number when the number of the written number is smaller than k. (1=<k<=n<=1000000).

 

Source

The 36th ACM/ICPC Asia Regional
Dalian Site —— Online Contest

 

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题目大意：

给出一系列数， 问第 k 大是多少。。。

思路：

是不是一万种处理方法，不过最近在看 sbt  就用 SBT 写的。

我感觉 SBT 这种数据结构很厉害啊   比伸展树  要厉害的多    简单粗暴  还快 。

实测  printf 比 cout 快 100多 ms

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=200010;
int N,M;
struct SBT
{
int left,right,size,key;
void init(int val)
{
left=right=0;
size=1;
key=val;
}
} tree[maxn];
int tot,root;
void left_rotate(int &x)
{
int y=tree[x].right;
tree[x].right=tree[y].left;
tree[y].left=x;
tree[y].size=tree[x].size;
tree[x].size=tree[tree[x].left].size+tree[tree[x].right].size+1;
x=y;
}
void right_rotate(int &x)
{
int y=tree[x].left;
tree[x].left=tree[y].right;
tree[y].right=x;
tree[y].size=tree[x].size;
tree[x].size=tree[tree[x].left].size+tree[tree[x].right].size+1;
x=y;
}
void maintain(int &x,int flag)
{
if(!flag)
{
if(tree[tree[tree[x].left].left].size>tree[tree[x].right].size)
right_rotate(x);
else if(tree[tree[tree[x].left].right].size>tree[tree[x].right].size)
left_rotate(tree[x].left),right_rotate(x);
else return;
}
else
{
if(tree[tree[tree[x].right].right].size>tree[tree[x].left].size)
left_rotate(x);

4000
else if(tree[tree[tree[x].right].left].size>tree[tree[x].left].size)
right_rotate(tree[x].right),left_rotate(x);
else return;
}
maintain(tree[x].left,0);
maintain(tree[x].right,1);
maintain(x,0);
maintain(x,1);
}
//插入值为key的节点
void insert(int &x,int key)
{
if(x==0)
{
x=++tot;
tree[x].init(key);
}
else
{
tree[x].size++;
if(key<tree[x].key)
insert(tree[x].left,key);
else
insert(tree[x].right,key);
maintain(x,key>=tree[x].key);
}
}

int del(int &x,int key)
{
if(!x)return 0;
int res=0;
tree[x].size--;
if(key==tree[x].key||(key<tree[x].key&&tree[x].left==0)||
(key>tree[x].key&&tree[x].right==0))
{
if(tree[x].left&&tree[x].right)
{
int p=del(tree[x].left,key+1);
tree[x].key=tree[p].key;
res=p;
}
else
{
int p=x;
x=tree[x].left+tree[x].right;
res=p;
}
}
else
res=del(key<tree[x].key?tree[x].left:tree[x].right,key);
maintain(x,key<tree[x].key);
return res;
}

void Delete(int &t,int delay)
{
if(t==0)return;
if(tree[t].key+delay<M)
{
t=tree[t].right;
Delete(t,delay);
}
else
{
Delete(tree[t].left,delay);
tree[t].size=tree[tree[t].left].size+tree[tree[t].right].size+1;
}
}
int find(int &t,int k)
{
if(k<=tree[tree[t].right].size)
return find(tree[t].right,k);
else if(k>tree[tree[t].right].size+1)
return find(tree[t].left,k-tree[tree[t].right].size-1);
return tree[t].key;
}
int main()
{
char op[10];
int x;
int tt;
while(scanf("%d%d",&N,&x)!=EOF)
{
int delay=0,ans=0;
root=tot=0;
while(N--)
{
scanf("%s",op);
if(op[0]=='I')
{
scanf("%d",&tt);
insert(root,tt);
}
else
{
printf("%d\n",find(root,x));
}
}
}
return 0;
}