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Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

Codeforces Round #616 (Div. 2) C. Mind Control 博弈论 枚举

C. Mind Control

time limit per test1 second
memory limit per test256 megabytes

You and your n−1 friends have found an array of integers a1,a2,…,an. You have decided to share it in the following way: All n of you stand in a line in a particular order. Each minute, the person at the front of the line chooses either the first or the last element of the array, removes it, and keeps it for himself. He then gets out of line, and the next person in line continues the process.

You are standing in the m-th position in the line. Before the process starts, you may choose up to k different people in the line, and persuade them to always take either the first or the last element in the array on their turn (for each person his own choice, not necessarily equal for all people), no matter what the elements themselves are. Once the process starts, you cannot persuade any more people, and you cannot change the choices for the people you already persuaded.

Suppose that you're doing your choices optimally. What is the greatest integer x such that, no matter what are the choices of the friends you didn't choose to control, the element you will take from the array will be greater than or equal to x?

Please note that the friends you don't control may do their choice arbitrarily, and they will not necessarily take the biggest element available.

Input

The input consists of multiple test cases. The first line contains a single integer t (1≤t≤1000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains three space-separated integers n, m and k (1≤m≤n≤3500, 0≤k≤n−1) — the number of elements in the array, your position in line and the number of people whose choices you can fix.

The second line of each test case contains n positive integers a1,a2,…,an (1≤ai≤109) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 3500.

Output

For each test case, print the largest integer x such that you can guarantee to obtain at least x.

Example

input
4
6 4 2
2 9 2 3 8 5
4 4 1
2 13 60 4
4 1 3
1 2 2 1
2 2 0
1 2
output
8
4
1
1

Note

In the first test case, an optimal strategy is to force the first person to take the last element and the second person to take the first element.

the first person will take the last element (5) because he or she was forced by you to take the last element. After this turn the remaining array will be [2,9,2,3,8];
the second person will take the first element (2) because he or she was forced by you to take the first element. After this turn the remaining array will be [9,2,3,8];
if the third person will choose to take the first element (9), at your turn the remaining array will be [2,3,8] and you will take 8 (the last element);
if the third person will choose to take the last element (8), at your turn the remaining array will be [9,2,3] and you will take 9 (the first element).
Thus, this strategy guarantees to end up with at least 8. We can prove that there is no strategy that guarantees to end up with at least 9. Hence, the answer is 8.

In the second test case, an optimal strategy is to force the first person to take the first element. Then, in the worst case, both the second and the third person will take the first element: you will end up with 4.

题意

现在有n个人,n个物品,你站在第m个位置,你可以在游戏的一开始控制k个人使得这k个人只能选择前面,或者只能选择后面的物品。

游戏的每一轮,人们按照顺序去拿物品,要么只能拿最前面的,要么只能拿最后面的,问你最少能拿到多大的价值的物品。

题解

视频题解:https://www.bilibili.com/video/av86529667/

首先控制的k个人,一定是前k个人。

我们枚举这k个人拿前面i个,那么后面就会有k-i个人,这个是由我控制的。

然后还剩下m-k-1个人,我们枚举前面有j个人,那么后面就会有m-k-1-j个人,这个是由对方控制的。

最后我从前面的一个物体,和后面的一个物体进行二选一,选择最大的,这个是由我控制的。

由我控制的部分取max,由别人控制的部分取min,暴力枚举就可以了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 4005;
int n,m,k,a[maxn];
void solve(){
    cin>>n>>m>>k;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    k=min(m-1,k);
    int ans = 0;
    for(int i=0;i<=k;i++){
        int tmp=1e9;
        for(int j=0;j<m-k;j++){
            tmp=min(tmp,max(a[i+j],a[n-1-(k-i)-(m-k-j-1)]));
        }
        ans=max(ans,tmp);
    }
    cout<<ans<<endl;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        solve();
    }
}
posted @ 2020-02-03 19:23  qscqesze  阅读(268)  评论(4编辑  收藏