HDOJ -- 4632 区间DP

Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 1977    Accepted Submission(s): 822

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

4 a aaaaa goodafternooneveryone welcometoooxxourproblems

 

Sample Output

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

 

 思路：设dp[i][j]表示区间[i,j]的回文串的个数，那么有dp[i][j] = dp[j+1][i] + dp[j][i-1] - dp[j+1][i-1]，如果str[i] == str[j]，那么dp[i][j]还要加上dp[j+1][i-1] + 1;

#include<iostream>
#include<cstdio>
#include<cstring>
#define MAX 1010
#define MOD 10007
using namespace std;
int dp[MAX][MAX];
char str[MAX];
int main(){
    int T, cnt = 0;
    //freopen("in.c", "r", stdin);
    scanf("%d", &T);
    while(T--){
        memset(str, 0, sizeof(str));
        memset(dp, 0, sizeof(dp));
        scanf("%s", str);
        int len = strlen(str);
        for(int i = 0; i< len;i ++) dp[i][i] = 1;
        for(int i = 0;i < len;i ++){
            for(int j = i-1;j >= 0;j --){
                dp[j][i] = (dp[j][i-1] + dp[j+1][i] - dp[j+1][i-1] + MOD)%MOD;
                if(str[i] == str[j]) dp[j][i] = (dp[j][i] + dp[j+1][i-1] + 1 + MOD)%MOD;
            }
        }
        printf("Case %d: %d
", ++cnt, dp[0][len-1]);
    }
    return 0;
}