ZOJ 2744 Palindromes（动态规划）

Palindromes

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDEDCBA" is a palindrome because it is the same when the string is read from left to right as when the string is read from right to left.

Now give you a string S, you should count how many palindromes in any consecutive substring of S.

Input

There are several test cases in the input. Each case contains a non-empty string which has no more than 5000 characters.

Proceed to the end of file.

Output

A single line with the number of palindrome substrings for each case.

Sample Input

aba
aa

Sample Output

4
3

 

做了一下午的动态规划，才发现懂得只有那么一点点，大部分还是百度出来才能理解状态转移方程，这一道，算是自己思考的最多的一道吧

 令d[i][j]表示这一串字符从第i个到第j个字符组成的串是否是回文。

如果 i 到 j 组成的串字符超过了3个，则必须满足 i-1 到j-1 个字符组成的串是回文，它才有可能是回文

初始条件是：数组所有数据为0

# include<stdio.h>
# include<string.h>
# define maxn 5001
char s[maxn];
bool dp[maxn][maxn];

int main()
{
    int len,ans,i,j,k;
    while(scanf("%s",s)!=EOF)
    {
        len=strlen(s);
        ans=len;
        memset(dp,0,sizeof(dp));
        for(k=1;k<len;k++)
        {
            for(i=0;i<len-k;i++)
            {
                j=i+k;
                dp[i][j]=0;
                if(s[i]==s[j])
                {
                    if(i+1 <j-1)
                    {
                        if(dp[i+1][j-1])
                        {
                            dp[i][j]=1;
                            ans++;
                        }
                    }
                    else
                    {
                        dp[i][j]=1;
                        ans++;
                    }
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}