AGC：D - ABS（博弈）

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本文介绍了一个涉及策略选择的游戏问题，两名玩家通过轮流抽取卡片并替换手中的牌来比拼得分，目标是最小化或最大化两者的分数差值。文章提供了解决此问题的一种思路及代码实现。

D - ABS

Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

We have a deck consisting of N cards. Each card has an integer written on it. The integer on the i-th card from the top is ai.

Two people X and Y will play a game using this deck. Initially, X has a card with Z written on it in his hand, and Y has a card with W written on it in his hand. Then, starting from X, they will alternately perform the following action:

Draw some number of cards from the top of the deck. Then, discard the card in his hand and keep the last drawn card instead. Here, at least one card must be drawn.

The game ends when there is no more card in the deck. The score of the game is the absolute difference of the integers written on the cards in the two players' hand.

X will play the game so that the score will be maximized, and Y will play the game so that the score will be minimized. What will be the score of the game?

Constraints

All input values are integers.
1≤N≤2000
1≤Z,W,ai≤109

Input

Input is given from Standard Input in the following format:

N Z W
a1 a2 … aN

Output

Print the score.

Sample Input 1

Copy

3 100 100
10 1000 100

Sample Output 1

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If X draws two cards first, Y will draw the last card, and the score will be |1000−100|=900.

Sample Input 2

Copy

3 100 1000
10 100 100

Sample Output 2

Copy

If X draws all the cards first, the score will be |1000−100|=900.

Sample Input 3

Copy

5 1 1
1 1 1 1 1

Sample Output 3

Copy

Sample Input 4

Copy

1 1 1
1000000000

Sample Output 4

Copy

999999999

题意：两个人初始手上分别有Z，W两个牌，一副N个牌在桌面上，每轮取任意>0数目的牌，将手上的牌替换成取得牌最下面那张，然后剩余的扔掉，双方轮流操作直至无牌。结果是两人手上牌的差值K，先手令K尽量大，后手令K尽量小，问结果是什么。

思路：答案是max(a[n]-a[n-1], a[n]-w)，这种题的思路我一直都比较模糊，记下我目前的想法，因为先手要最大化差值，开局他可以取a[n-1]剩下a[n]，或者取a[n]直接结束游戏，这两种情况是先手100%能达到的。如果先手不这样做，他取得牌点数必须>=a[n-1]，否则后手会直接拿掉a[n]，这样先手就违背他的最大化差值的使命了。然后同理后手就取<=a[n-1]的牌，否则将违背他的使命，这样最终肯定是分别拿到a[n]和a[n-1]的。

# include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a[2003];
int main()
{
    int n;
    LL z, w;
    scanf("%d%lld%lld",&n,&z,&w);
    for(int i=1; i<=n; ++i) scanf("%d",&a[i]);
    if(n == 1) printf("%lld\n",llabs(a[1]-w));
    else
        printf("%lld\n",max(llabs(a[n-1]-a[n]), llabs(a[n]-w)));
    return 0;
}