Again Palindromes - UVa 10617 dp

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探讨了如何通过删除字符使普通字符串变为回文串的方法总数，并提供了详细的DP算法实现思路及代码。

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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, TOTand MADAM 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

BAOBAB

AAAA

ABA

题意：删除字符串中任意数量的字符，使得其成为一个回文串，问这种删法一共有多少种，不删也算一种。

思路：还是从边上往中间递推if(s[l]==s[r])　　dp[l][r]=dp[l+1][r]+dp[l][r-1]+1;　　else dp[l][r]=dp[l+1][r]+dp[l][r-1]-dp[l+1][r-1];

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
long long dp[100][100],;
char s[110];
void dfs(int l,int r)
{ if(dp[l][r]>=0)
   return;
  if(l>r)
  { dp[l][r]=0;
    return;
  }
  dfs(l+1,r);
  dfs(l,r-1);
  dfs(l+1,r-1);
  if(s[l]==s[r])
   dp[l][r]=dp[l+1][r]+dp[l][r-1]+1;
  else
   dp[l][r]=dp[l+1][r]+dp[l][r-1]-dp[l+1][r-1];
}
int main()
{ int t,n,m,i,j,k,len;
  scanf("%d",&t);
  while(t--)
  { scanf("%s",s+1);
    len=strlen(s+1);
    memset(dp,-1,sizeof(dp));
    for(i=1;i<=len;i++)
     dp[i][i]=1;
    dfs(1,len);
    printf("%lld\n",dp[1][len]);
  }
}