UVa 10617 Again Palindrome 字符串dp

Problem I

Again Palindromes

Input: Standard Input

Output: Standard Output

Time Limit: 2 Seconds

 

A palindorme is a sequence of one or more characters that reads the same from the left as it does from the right. For example, Z, TOT andMADAM are palindromes, but ADAM is not.

Given a sequence S of N capital latin letters. How many ways can one score out a few symbols (maybe 0) that the rest of sequence become a palidrome. Varints that are only  different by an order of scoring out should be considered the same.

Input

The input file contains several test cases (less than 15). The first line contains an integer T that indicates how many test cases are to follow.

Each of the T lines contains a sequence S (1≤N≤60). So actually each of these lines is a test case.

 

Output

For each test case output in a single line an integer – the number of ways.

Sample Input                             Output for Sample Input

3 BAOBAB AAAA ABA

22 15 5

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Russian Olympic Camp

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忽然发现我的字符串dp弱成渣啊。。。

f[i][j]表示从i到j的回文串个数。

当i!=j时，f[i][j]=f[i+1][j]+f[i][j-1]-f[i+1][j-1];从i到j的回文数=删除左边的字符后的回文数+删除右边字符后的回文数-重复部分

当i==j时，f[i][j]=f[i+1][j]+f[i][j-1]-f[i+1][j-1]+f[i+1][j-1]+1;从i到j的回文数=删除左边的字符后的回文数+删除右边字符后的回文数-重复部分+保留ij后的回文数+只留下ij的回文数即1

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int T;
char s[111];
long long f[111][111];
int n;

long long dfs(int l,int r)
{
    if (f[l][r]!=-1) return f[l][r];
    if (r-l<0) return 0;
    if (s[l]==s[r])
    {
        f[l][r]=dfs(l+1,r)+dfs(l,r-1)+1;
    }
    else
    {
        f[l][r]=dfs(l+1,r)+dfs(l,r-1)-dfs(l+1,r-1);
    }
    return f[l][r];
}

int main()
{
    cin>>T;
    while (T--)
    {
        memset(f,-1,sizeof(f));
        cin>>(s+1);
        n=strlen(s+1);
        long long ans=dfs(1,n);
        cout<<ans<<endl;
    }
    return 0;
}