HDOJ 4745 Two Rabbits DP
Two Rabbits
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 944 Accepted Submission(s): 496
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.
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.
1
1
4
1 1 2 1
6
2 1 1 2 1 3
0
1
4
5HintFor 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.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; const int maxn=2200; int n;
int a[maxn];
int dp[maxn][maxn]; int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
a[n+i]=a[i];
}
memset(dp,0,sizeof(dp));
n+=n;
for(int i=0;i<n;i++)
{
dp[i][i]=1;
}
for(int len=2;len<n;len++)
{
for(int i=0;i+len-1<n;i++)
{
int j=i+len-1;
dp[i][j]=max(dp[i][j],max(dp[i+1][j],dp[i][j-1]));
if(a[i]==a[j])
{
dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
}
}
}
int ans=1;
for(int i=0;i+n/2-1<n;i++)
{
ans=max(ans,dp[i][i+n/2-1]);
}
for(int i=0;i+n/2-2<n;i++)
{
ans=max(ans,dp[i][i+n/2-2]+1);
}
printf("%d\n",ans);
}
return 0;
}
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