String Game【ZSTUOJ--4212】

Description

Alice and Bob are playing the following game with strings of letters.

Before the game begins, an initial string and a target string are decided. The initial string is at least as long as the target string. Then, Alice and Bob take turns, starting with the initial
string. Bob goes first. In each turn, the current player removes either the first or the last letter of the current string. Once the length of the current string becomes equal to the length of the target string, the game stops. If the string at the end of
the game is equal to the target string, Alice wins the game; otherwise Bob wins.

Determine who will win the game if both players are playing optimally.

题意：给你两串字符串a和b，Alice和Bob轮流去掉第一个字符串a的头字母或尾字母，问经过去掉头尾字母后能否得到字符串b，若能得到则Alice赢；否则Bob赢，Bob第一个开始，输出赢家。

分析：刚开始以为是道博弈题，但是由于弱太渣，没推出来233，直到比赛结束看了题解才会这道题

。题解上的这道题的解不是用的博弈，而是直接用的暴力枚举，当然不是简单的纯暴力，是有规律的

。据题解知（当然你最好自己推一下，可以发现规律哦

），当b在a字符串的中间时，Alice肯定会赢，因为不管Bob去头还是去尾，Alice都可以去另一头以保证中间的永远留着，这是b字符串在a字符串中出现一次的情况下，如果b字符串在a字符串中出现了两次，则当这两个字符串在a中关于中间对称且两个字符串位置差不超过2时，不管Bob怎么去，Alice都可以与Bob同方向去掉字母以保证让b字符串留下。具体步骤请见代码。

Input

Each
test case starts with N,
the number of inputs to process. Each input consists of one line, which contains the initial

string,
followed by a space, followed by the target string. Each string consists of only lowercase letters. The total input

length
will be less than 500000 characters.

Output

For
each input, output the winner, which will either be Alice or Bob.

Sample Input

5
aba b
bab b
aaab aab
xyz mnk
xyz xyz

Sample Output

Alice
Alice
Bob
Bob
Alice

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char a[500001],b[500001];
int settle()
{
int len1=strlen(a);
int len2=strlen(b);
if(len1<len2)
return 0;
int pos=(len1-len2)/2;        //b字符串在a字符串中可能出现的中间位置
int flag;
if(len1%2==len2%2)            //a字符串与b字符串长度同奇偶时
{
flag=1;
for(int i=0; i<len2; i++) //b字符串在a字符串中只出现一次
if(b[i]!=a[pos+i])
{
flag=0;
break;
}
if(flag)
return 1;
flag=1;
for(int i=0; i<len2; i++)  //b字符串在a字符串中出现两次
if(b[i]!=a[pos+i+1])
{
flag=0;
break;
}
for(int i=0; i<len2; i++)
if(b[i]!=a[pos+i-1])
{
flag=0;
break;
}
}
else                            //a字符串与b字符串长度奇偶不同时
{
flag=1;
for(int i=0; i<len2; i++)
if(b[i]!=a[pos+i])
{
flag=0;
break;
}
for(int i=0; i<len2; i++)
if(b[i]!=a[pos+i+1])
{
flag=0;
break;
}
}
return flag;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%s %s",a,b);
printf("%s",settle()?"Alice\n":"Bob\n");
}
return 0;
}