HDU4745 Two Rabbits（区间dp）

Two Rabbits

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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.

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题意

两只兔子在n块围成环形的石头上跳跃，每块石头有一个重量值ai，一只顺时针跳一只逆时针跳，每个时刻要求两只兔子所在石头重量相同，兔子最多跳一圈（不回起点），问它们最多经过多少块石头.

分析

其实就是求一个环中，非连续最长回文子序列的长度。
dp[i][j] = max(dp[i + 1][j], max(dp[i][j - 1], (a[i] == a[j] ? dp[i + 1][j - 1] + 2 : 0)));
当然这只是求出一个序列的非连续最长回文子序列长度，而题目是环。
其实我们会发现可以把它当成一个链，然后切成两半，求出两边的回文长度，最大的和就是解。(这里不用考虑起点问题，因为两边的回文中点都可以做起点 )

CODE

代码其实很短

#include<cstdio>
#include<memory.h>
#define N 1005
#define FOR(i,a,b) for(int i=(a),i##_END_=(b);i<=i##_END_;i++)
int dp[N][N],a[N];
inline int  max(int x,int y) {return x>y?x:y;}
inline void Max(int&x,int y) {if(x<y)x=y;}
int main() {
    int n;
    while(scanf("%d",&n)&&n) {
        FOR(i,1,n)scanf("%d",&a[i]);
        memset(dp,0,sizeof dp);
        FOR(i,1,n)dp[i][i]=1;
        FOR(len,1,n-1)FOR(i,1,n-len) {
            int j=i+len;
            dp[i][j]=max(dp[i+1][j],max(dp[i][j-1],(a[i]==a[j]?dp[i+1][j-1]+2:0)));//状态转移
        }
        int ans=0;
        FOR(i,1,n)Max(ans,dp[1][i]+dp[i+1][n]);
        FOR(i,1,n)printf("dp[1][%d]=%d,dp[%d][%d]=%d\n",i,dp[1][i],i+1,n,dp[i+1][n]);
        printf("%d\n",ans);
    }
    return 0;
}

这是第一篇博客哦