山東省第六屆ACM大學生程式設計競賽-Lowest Unique Price(桶排序)

kewlgrl發表於2016-05-23
ACM程式設計排序

Lowest Unique Price

Time Limit: 1000ms   Memory limit: 65536K  有疑問？點這裡^_^

題目描述

Recently my buddies and I came across an idea! We want to build a website to sell things in a new way.
For each product, everyone could bid at a price, or cancel his previous bid, finally we sale the product to the one who offered the "lowest unique price". The lowest unique price is defined to be the lowest price that was called only once.
So we need a program to find the "lowest unique price", We'd like to write a program to process the customers' bids and answer the query of what's the current lowest unique price.
All what we need now is merely a programmer. We will give you an "Accepted" as long as you help us to write the program.
 

輸入

The first line of input contains an integer T, indicating the number of test cases (T ≤ 60).
Each test case begins with a integer N (1 ≤ N ≤ 200000) indicating the number of operations.
Next N lines each represents an operation.
There are three kinds of operations:
"b x": x (1 ≤ x ≤ 106) is an integer, this means a customer bids at price x.
"c x": a customer has canceled his bid at price x.
"q" : means "Query". You should print the current lowest unique price.
Our customers are honest, they won\'t cancel the price they didn't bid at.
 

輸出

 Please print the current lowest unique price for every query ("q"). Print "none" (without quotes) if there is no lowest unique price.

示例輸入

2 
3 
b 2 
b 2 
q 
12 
b 2 
b 2 
b 3 
b 3 
q 
b 4 
q 
c 4 
c 3 
q 
c 2 
q

示例輸出

none 
none 
4 
3 
2

提示

 

來源

題目意思：

b x 表示以x的價格參與競標；

c x 表示取消價格為n的競標；

q 表示查詢只有一人出價的最低價格。

 

解題思路：

試著用了一下桶排序，沒想到一下子就過了~

注意輸入輸出方式~純用cin/cout回超時~

 
#include <iostream>
#include <cstring>
#include <set>
using namespace std;
#define MAXN 200002
int a[MAXN];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        memset(a,0,sizeof(a));
        int n,i,j,x;
        cin>>n;
        for(i=0; i<n; ++i)
        {
            char c;
            cin>>c;
            if(c=='b')
            {
                cin>>x;
                ++a[x];
            }
            else if(c=='c')
            {
                cin>>x;
                --a[x];
            }
            else if(c=='q')
            {
                int flag=0;
                for(j=0; j<MAXN; ++j)
                    if(a[j]&&a[j]==1)
                    {
                        cout<<j<<endl;
                        flag=1;
                        break;
                    }
                if(flag==0) cout<<"none"<<endl;
            }
        }
    }
    return 0;
}
/*
2
3
b 2
b 2
q
12
b 2
b 2
b 3
b 3
q
b 4
q
c 4
c 3
q
c 2
q
*/

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