山东省第六届ACM大学生程序设计竞赛-Lowest Unique Price(桶排序)

Lowest Unique Price

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

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.

 

输入

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.

 

输出

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

示例输入

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

示例输出

none none 4 3 2

提示

 

来源

题目意思：

b x 表示以x的价格参与竞标；

c x 表示取消价格为n的竞标；

q 表示查询只有一人出价的最低价格。

 

解题思路：

试着用了一下桶排序，没想到一下子就过了~

注意输入输出方式~纯用cin/cout回超时~

 

#include <iostream>#include <cstring>#include <set>using namespace std;#define MAXN 200002int 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;}/*23b 2b 2q12b 2b 2b 3b 3qb 4qc 4c 3qc 2q*/