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hdu4745(区间DP)
Two Rabbits
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 921 Accepted Submission(s): 483
Problem 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.
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.
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
题意:RT
思路:这题可以简化为求环上的最长回文,为了方便操作直接将原序列复制一份到后面,然后在这个长度为2*n的序列上求一遍区间最长回文就可以了
DP[i][j]表示区间[i,j]的最长回文串,可以转移当a[i]==a[j],DP[i][j]=DP[i+1][j-1]+2
然后在所有长度范围为n的区间里找一个最大的即可
还有一种情况,也可以找区间长度为n-1的回文,但是答案还要加1,因为多出一个数是不影响兔子的最后一次跳的。。
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hdu4745(区间DP)
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