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): 558    Accepted Submission(s): 203

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

 

 

Source

2013 Multi-University Training Contest 4

 

 

Recommend

zhuyuanchen520

 

 第一次接触，

题意：

一个字符串，有多少个subsequence是回文串。

题解：

用dp[i][j]表示这一段里有多少个回文串，那首先dp[i][j]=dp[i+1][j]+dp[i][j-1]，但是dp[i+1][j]和dp[i][j-1]可能有公共部分，所以要减去dp[i+1][j-1]。

如果str[i]==str[j]的话，还要加上dp[i+1][j-1]+1。

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=1010;
const int mod=10007;

char str[N];
int dp[N][N];

int main(){

    //freopen("input.txt","r",stdin);

    int t,cases=0;
    scanf("%d",&t);
    while(t--){
        scanf("%s",str);
        int len=strlen(str);
        memset(dp,0,sizeof(dp));
        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\n",++cases,dp[0][len-1]);
    }
    return 0;
}

 

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=1010;
const int mod=10007;

char str[N];
int dp[N][N];

int DFS(int l,int r){
    if(l>r)
        return 0;
    if(l==r)
        return 1;
    if(dp[l][r]!=-1)
        return dp[l][r];
    dp[l][r]=(DFS(l+1,r)+DFS(l,r-1)-DFS(l+1,r-1)+mod)%mod;
    if(str[l]==str[r])
        dp[l][r]=(dp[l][r]+DFS(l+1,r-1)+1+mod)%mod;
    return dp[l][r];
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,cases=0;
    scanf("%d",&t);
    while(t--){
        scanf("%s",str+1);
        int len=strlen(str+1);
        memset(dp,-1,sizeof(dp));
        printf("Case %d: %d\n",++cases,DFS(1,len));
    }
    return 0;
}