传送门 The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, “101101” is a palindrome, while “0101” is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters ‘0’ and ‘1’. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
Choose any i (1≤i≤n), where s[i]= ‘0’ and change s[i] to ‘1’. Pay 1 dollar.
Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can’t reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, “01001” becomes “10010” after reversing.
The game ends when every character of string becomes ‘1’. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1≤t≤103). Then t test cases follow.
The first line of each test case contains a single integer n (1≤n≤103).
The second line of each test case contains the string s of length n, consisting of the characters ‘0’ and ‘1’. It is guaranteed that the string s contains at least one ‘0’.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
“ALICE”, if Alice will win the game,
“BOB”, if Bob will win the game,
“DRAW”, if the game ends in a draw.
思路：
先找到字符串s中0的个数，记为cnt，再找到使该字符串变为回文串的最少改变次数，记为b，然后cnt = cnt-b；
①
当cnt为奇数时，并且s不为回文串，则易得BOB的最优决策为尽快使s变为回文串，这样他才有与ALICE周旋的可能（即在s不为回文串时，ALICE总是可以选择翻转s），所以ALICE的最优决策为在s变为回文串之前一直翻转s串，这样在s串变为回文串时，BOB就比ALICE多花费了b，最后结合s为回文串时的情况进行特判就行。
②
当cnt为偶数时，BOB和ALICE的最优决策同上，有所差别的是，ALICE需要在最后一次能使s变为回文串的时候将对应的0变为1使得s成为回文串，这时候问题就变成了BOB先手的0的个数为偶数的回文串游戏，于是容易得出在这个游戏里，BOB始终比ALICE多花费2，又因为在s变为回文串的过程中，ALICE的最多花费为1，所以综上所述：当cnt为偶数并且s初始不是回文串时，始终是ALICE赢，否则BOB赢（即b=0时BOB赢）
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod = 998244353;
char s[1010];
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
scanf("%s",s+1);
int cnt = 0;
int a = 0,b = 0;
for(int i = 1; i <= n; i++)
{
if(s[i] == '0')cnt++;
}
int l = 0,r = 0;
for(int i = 1; i <= n/2; i++)
if(s[i] != s[n-i+1])
b++;
cnt -= b;
if(cnt % 2)
{
if(b == 1 && cnt == 1)
{
cout<<"DRAW"<<endl;
}
else if(b == 0)
{
if(cnt > 1)
cout<<"ALICE"<<endl;
else
cout<<"BOB"<<endl;
}
else
{
cout<<"ALICE"<<endl;
}
}
else
{
if(b == 0)
{
cout<<"BOB"<<endl;
}
else
{
cout<<"ALICE"<<endl;
}
}
}
}