hdu4745区间dp

B - Two Rabbits

Time Limit:5000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u

Submit  Status

Description

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.

 

Input

The 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.

 

Output

For 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：有一个环形的石头阵，每一块石头有一个权值，两只兔子任意选择一块石头相向而跳，规定两只兔子每次跳的石头权值相同，并且不可以回到出发点，相遇可以在同一块石头上，不规定跳的距离，即可以跳到规定方向任意远，问两只兔子可以跳的最远距离；

可以看出是在环上求最长回文串，样例2 1 1 2 1 3，长度是5，其实可以表示成区间[1-4]的回文串长度加上区间[5-6]最长回文串长度，结果ans=max（dp[1][i]+dp[i+1][n]，ans）;

ac代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
using namespace std;
const int maxn=1005;
int dp[maxn][maxn];
int d[maxn];
int max(int a,int b)
{
    if(a>b)
        return a;
return b;
}
int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
        {
             scanf("%d",&d[i]);
             dp[i][i]=1;
        }
        for(int i=n-1;i>=0;i--)
        {
            for(int j=i+1;j<n;j++)
            {
                if(d[i]==d[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=dp[0][n-1];
        for(int i=0;i<n-1;i++)
        ans=max(ans,dp[0][i]+dp[i+1][n-1]);

        printf("%d\n",ans);
    }
    //cout << "Hello world!" << endl;
    return 0;
}