ZCMU—B

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B - Lowest Unique Price

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

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.

Input

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.

Output

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

Sample Input

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

Sample Output

none none 4 3 2

【分析】

题意：给出很多命令

b表示加入一个价格，c表示删除一个价格，q表示询问当前最小价格

并且有个条件，当且只当某个价格x在队列中出现的次数为1的时候才被认为是可行的价格

set+判重...开考虑10^6所以不开一般数组用map...当然用一般数组也不会炸内存..但是memset的初始化速度可能会导致tle..

【代码】

#include <iostream>#include <set>#include <cstdio>#include <map>#include <algorithm>#include <cstring>using namespace std;char s[10];map<int,int>vis;set<int>a; int main(){int pp;scanf("%d",&pp);while (pp--){vis.clear();a.clear();int n,x;scanf("%d",&n);for (int i=0;i<n;i++){scanf("%s",s);if (s[0]=='b'){scanf("%d",&x);vis[x]++;if (vis[x]==1) a.insert(x);else a.erase(x);}elseif (s[0]=='c'){scanf("%d",&x);vis[x]--;if (vis[x]==1) a.insert(x);if (vis[x]==0) a.erase(x);}else{if (a.empty()) printf("none\n");elseprintf("%d\n",*a.begin());}}}}

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