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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11824 | Accepted: 5385 |
Description
The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest, equipped with a modern computing environment provided by IBM Romania, and using modern information technologies. As usual, each client of the bank is identified by a positive integer K and, upon arriving to the bank for some services, he or she receives a positive integer priority P. One of the inventions of the young managers of the bank shocked the software engineer of the serving system. They proposed to break the tradition by sometimes calling the serving desk with the lowest priority instead of that with the highest priority. Thus, the system will receive the following types of request:
| 0 | The system needs to stop serving |
| 1 K P | Add client K to the waiting list with priority P |
| 2 | Serve the client with the highest priority and drop him or her from the waiting list |
| 3 | Serve the client with the lowest priority and drop him or her from the waiting list |
Your task is to help the software engineer of the bank by writing a program to implement the requested serving policy.
Input
Each line of the input contains one of the possible requests; only the last line contains the stop-request (code 0). You may assume that when there is a request to include a new client in the list (code 1), there is no other request in the list of the same client or with the same priority. An identifier K is always less than 106, and a priority P is less than 107. The client may arrive for being served multiple times, and each time may obtain a different priority.
Output
For each request with code 2 or 3, the program has to print, in a separate line of the standard output, the identifier of the served client. If the request arrives when the waiting list is empty, then the program prints zero (0) to the output.
Sample Input
2 1 20 14 1 30 3 2 1 10 99 3 2 2 0
Sample Output
0 20 30 10 0
Source
杭电上的原题，之前用map做的，学了SBT之后再poj用SBT做了一下
注释部分是找前驱版本的删除，，看上去貌似稍慢了一些
ac代码
#include<stdio.h>
#include<string.h>
//#pragma comment(linker, "/STACK:102400000,102400000")
struct s
{
int key,left,right,size,num;
}tree[1000100];
int top,root,keynum;
void left_rot(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_rot(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,bool flag)//维护SBT状态
{
if(flag==false)
{
if(tree[tree[tree[x].left].left].size>tree[tree[x].right].size)
right_rot(x);
else
if(tree[tree[tree[x].left].right].size>tree[tree[x].right].size)
{
left_rot(tree[x].left);
right_rot(x);
}
else
return;
}
else
{
if(tree[tree[tree[x].right].right].size>tree[tree[x].left].size)
left_rot(x);
else
if(tree[tree[tree[x].right].left].size>tree[tree[x].left].size)
{
right_rot(tree[x].right);
left_rot(x);
}
else
return;
}
maintain(tree[x].left,false);
maintain(tree[x].right,true);
maintain(x,true);
maintain(x,false);
}
void insert(int &x,int key,int num)
{
if(x==0)
{
x=++top;
tree[x].left=0;
tree[x].right=0;
tree[x].size=1;
tree[x].key=key;
tree[x].num=num;
}
else
{
tree[x].size++;
if(key<tree[x].key)
insert(tree[x].left,key,num);
else
insert(tree[x].right,key,num);
maintain(x,key>=tree[x].key);
}
}
int getmin(int x)
{
while(tree[x].left)
x=tree[x].left;
//keynum=tree[x].key;
return x;
}
int getmax(int x)
{
while(tree[x].right)
x=tree[x].right;
keynum=tree[x].key;
return x;
}
/*int remove(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=remove(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=remove(key<tree[x].key?tree[x].left:tree[x].right,key);
maintain(x,key<tree[x].key);
return res;
} */
int remove(int &x,int key)
{
tree[x].size--;
if(key>tree[x].key)
remove(tree[x].right,key);
else
if(key<tree[x].key)
remove(tree[x].left,key);
else
if(tree[x].left!=0&&tree[x].right==0)
{
int temp=x;
x=tree[x].left;
return temp;
}
else
if(!tree[x].left&&tree[x].right!=0)
{
int temp=x;
x=tree[x].right;
return temp;
}
else
if(!tree[x].left&&!tree[x].right)
{
int temp=x;
x=0;
return temp;
}
else
{
int temp=tree[x].right;
while(tree[temp].left)
temp=tree[temp].left;
tree[x].key=tree[temp].key;
remove(tree[x].right,tree[temp].key);
}
}
int main()
{
int n;
root=top=0;
while(scanf("%d",&n)!=EOF,n)
{
if(n==1)
{
int key,num;
scanf("%d%d",&num,&key);
insert(root,key,num);
}
else
if(n==2)
{
int x=getmax(root);
printf("%d\n",tree[x].num);
remove(root,tree[x].key);
}
else
if(n==3)
{
int x=getmin(root);
printf("%d\n",tree[x].num);
remove(root,tree[x].key);
}
}
}
ac代码
Problem: 3481 User: kxh1995
Memory: 196K Time: 297MS
Language: C++ Result: Accepted
Source Code
#include<stdio.h>
#include<string.h>
#include<map>
#include<iostream>
using namespace std;
map<int,int>m;
int main()
{
int n;
while(scanf("%d",&n)!=EOF,n)
{
int k,p;
if(n==1)
{
scanf("%d%d",&k,&p);
m.insert(pair<int,int>(p,k));
}
else
if(n==2)
{
if(m.empty())
{
printf("0\n");
}
else
{
printf("%d\n",m.rbegin()->second);
m.erase(m.rbegin()->first);
}
}
else
if(n==3)
{
if(m.empty())
{
printf("0\n");
}
else
{
printf("%d\n",m.begin()->second);
m.erase(m.begin()->first);
}
}
}
}
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POJ 题目3481 Double Queue（SBT ro map）
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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/47622131
