区间dp

Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny afternoon, they planned to play a game with some stones. There were n stones on the ground and they were arranged as a clockwise ring. That is to say, the first stone was adjacent to the second stone and the n-th stone, and the second stone is adjacent to the first stone and the third stone, and so on. The weight of the i-th stone is ai.

The rabbits jumped from one stone to another. Tom always jumped clockwise, and Jerry always jumped anticlockwise.

At the beginning, the rabbits both choose a stone and stand on it. Then at each turn, Tom should choose a stone which have not been stepped by itself and then jumped to it, and Jerry should do the same thing as Tom, but the jumping direction is anti-clockwise.

For some unknown reason, at any time , the weight of the two stones on which the two rabbits stood should be equal. Besides, any rabbit couldn't jump over a stone which have been stepped by itself. In other words, if the Tom had stood on the second stone, it cannot jump from the first stone to the third stone or from the n-the stone to the 4-th stone.

Please note that during the whole process, it was OK for the two rabbits to stand on a same stone at the same time.

Now they want to find out the maximum turns they can play if they follow the optimal strategy.

InputThe input contains at most 20 test cases.
For each test cases, the first line contains a integer n denoting the number of stones.
The next line contains n integers separated by space, and the i-th integer ai denotes the weight of the i-th stone.(1 <= n <= 1000, 1 <= ai <= 1000)
The input ends with n = 0.OutputFor each test case, print a integer denoting the maximum turns.Sample Input

1
1
4
1 1 2 1
6
2 1 1 2 1 3
0

Sample Output

1
4
5

        
 

Hint

For the second case, the path of the Tom is 1, 2, 3, 4, and the path of Jerry is 1, 4, 3, 2.
For the third case, the path of Tom is 1,2,3,4,5 and the path of Jerry is 4,3,2,1,5.

        
 

题意 ： 给你一个环形的串，上面写有一些数字，两只兔子向着相反的方向去跳，任意选择起点，并且要求每一轮两只兔子所占的数字是相同，问最多能进行几轮？

思路分析：其实就是让求一个最长回文的串，对于环的话，我们可以将长度拉直并延伸一倍，对新的串求区间内的回文，比较好写，然后就是在 dp求完任意一个区间内的最长回文串后，我们需要做的就是枚举下区间的头，判断一下长度为 n的区间内的最长回文串的长度最长是多少

代码示例：

int n;
int pre[2005];
int dp[2005][2005];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    while(~scanf("%d", &n) && n){
        for(int i = 1; i <= n; i++){
            scanf("%d", &pre[i]);        
            pre[n+i] = pre[i];
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= 2*n; i++) dp[i][i] = 1;
        
        for(int len = 2; len <= n; len++){
            for(int i = 1; i <= 2*n; i++){
                int j = i+len-1;
                if (j > 2*n) break;
                if (pre[i] == pre[j]) dp[i][j] = dp[i+1][j-1]+2;
                else dp[i][j] = max(dp[i+1][j], dp[i][j-1]);
            }
        }
        int ans = 0;
        for(int i = 1; i <= n; i++) ans = max(ans, dp[i][i+n-1]); // 不共起点
        for(int i = 1; i <= n; i++) ans = max(ans, dp[i][i+n-2]+1); // 共起点
        printf("%d
", ans);
    }
    return 0;
}