http://acm.hdu.edu.cn/showproblem.php?pid=4745
题意：
有两只兔子Tom Jerry， 他们在一个用石头围城的环形的路上跳， Tom只能顺时针跳，Jerry只能逆时针跳， 要求在跳的过程中他们所在石头的权值必须相同，而且只能单向跳，中间不能有已经跳过的石头。
思路：
模型就是求环上的最长回文串，我们只要将原串倍增，然后每个长度为n的子串的最长回文串就是我们要求的。区间DP一下就好了， 注意要考虑起点终点是统一点的情况特殊。
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x) (x) << 1
#define R(x) (x) << 1 | 1
#define MID(l, r) (l + r) >> 1
#define Min(x, y) (x) < (y) ? (x) : (y)
#define Max(x, y) (x) < (y) ? (y) : (x)
#define E(x) (1 << (x))
#define iabs(x) (x) < 0 ? -(x) : (x)
#define OUT(x) printf("%I64d\n", x)
#define keyTree (chd[chd[root][1]][0])
#define Read() freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout); #define M 107
#define N 1007 using namespace std; int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1}; const int inf = 0x7f7f7f7f;
const int mod = 1000000007;
const double eps = 1e-8;
const int R = 100007; int dp[2*N][2*N];
int a[2*N];
int n; int main()
{
while (~scanf("%d",&n))
{
if (!n) break;
CL(dp,0);
for (int i = 1; i <= n; ++i)
{
scanf("%d",&a[i]);
a[i + n] = a[i];
dp[i][i] = dp[i + n][i + n] = 1;
}
//没句串的长度
for (int i = 2; i <= n; ++i)
{
for (int j = 1; j + i - 1 <= 2*n; ++j)
{
int k = j + i - 1;
if (a[j] == a[k]) dp[j][k] = dp[j + 1][k - 1] + 2;
else
{
dp[j][k] = max(dp[j][k], max(dp[j + 1][k], dp[j][k - 1]));
}
}
}
int ans = 0;
for (int i = 1; i <= n; ++i) ans = max(ans, dp[i][i + n - 1]);
//起点终点是同一点的情况
for (int i = 1; i <= n; ++i) ans = max(ans, dp[i][i + n - 2] + 1);
printf("%d\n",ans);
}
return 0;
}