HDU 4632 Palindrome subsequence（区间DP）

Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 3309    Accepted Submission(s): 1396

Problem Description

In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a subsequence of <A, B, C, D, E, F>.
(http://en.wikipedia.org/wiki/Subsequence)

Given a string S, your task is to find out how many different subsequence of S is palindrome. Note that for any two subsequence X = <S_x1, S_x2, ..., S_xk> and Y = <S_y1, S_y2, ..., S_yk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if S_xi = S_yi. Also two subsequences with different length should be considered different.

 

Input

The first line contains only one integer T (T<=50), which is the number of test cases. Each test case contains a string S, the length of S is not greater than 1000 and only contains lowercase letters.

 

Output

For each test case, output the case number first, then output the number of different subsequence of the given string, the answer should be module 10007.

 

Sample Input

4aaaaaagoodafternooneveryonewelcometoooxxourproblems

 

Sample Output

Case 1: 1Case 2: 31Case 3: 421Case 4: 960

 

Source

2013 Multi-University Training Contest 4 

 

题意：

输出字符串中子序列是回文的数量%10007

POINT：

对每个区间[l,r]=[l+1,r]+[l,r-1]-[l+r,r-1]。

每个区间如果s[l]==s[r]则在加上[l+r,r-1]+1。

#include <stdio.h>#include <iostream>#include <vector>#include <algorithm>#include <queue>#include <string.h>using namespace std;#define LL long longconst int maxn = 1111;int dp[maxn][maxn];int main(){    int T;    cin>>T;    int p=0;    while(T--){        memset(dp,0,sizeof dp);        char s[maxn];        scanf("%s",s+1);        s[0]=1;        int l=strlen(s)-1;        for(int i=1;i<=l;i++) dp[i][i]=1;        for(int len=2;len<=l;len++){            for(int i=1;i<=l-len+1;i++){                int j=i+len-1;                if(s[i]==s[j]) (dp[i][j]+=dp[i+1][j-1]+1)%=10007;                (dp[i][j]+=(10007+dp[i][j-1]+dp[i+1][j]-dp[i+1][j-1]))%=10007;            }        }        printf("Case %d: ",++p);        printf("%d\n",dp[1][l]%10007);    }    return 0;}