hdu 4745 Two Rabbits 区间DP

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
", 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
",ans);
    }
    return 0;
}