685

# 题目

## Description

Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.

Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.

The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.

"1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
"2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?

## Input

The first line contains three integers n, k and q (1 ≤ n, q ≤ 150000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.

The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.

The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.

It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the

## Output won't be empty.

Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.

## Sample Input

4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3

>>

NO
YES
NO
YES
YES

>>

### Input

6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3

NO
YES
NO
YES

# 代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！

*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
using namespace std;

const int maxn = 150005;

int weight[maxn];

struct Node {
int n;
Node(int a) {
n = a;
}
bool operator < (const Node &rhs)const {
return weight[n] > weight[rhs.n];
}
};

priority_queue<Node> Q;
bool flag[maxn];

void init() {
memset(flag,false,sizeof(flag));
while(!Q.empty())
Q.pop();
}

bool Do() {
int n,k,q;
if(scanf("%d%d%d",&n,&k,&q)==EOF)
return false;
init();
for(int i = 1;i <= n;i++)
scanf("%d",&weight[i]);

for(int i = 0;i < q;i++) {
int com;
scanf("%d",&com);
if(com == 1) {
int t;
scanf("%d",&t);
flag[t] = true;
Q.push(Node(t));

while((int)Q.size() > k) {
flag[Q.top().n] = false;
Q.pop();
}

} else {
int t;
scanf("%d",&t);

if(flag[t])
printf("YES\n");
else
printf("NO\n");
}
}

//printf("\n");
return true;
}

int main() {
while(Do());
return 0;
}

• 点击查看/关闭被识别为广告的评论