hdu 4632(区间dp)

Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2858    Accepted Submission(s): 1168

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

 

 

题意：找出一个串中所有"回文串"的个数。这个回文串的意思是：所有子串（包裹不相邻的）只要满足回文串的性质都属于回文串.

比如 acba 里面总共有 5个回文串 a,c,b,a,aa

 

题解：区间DP，对于 i - j 区间，它满足条件的子串数量为 dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1] （子问题之和-重复子问题）

如果str[i] ==str[j] 的话 ，那么在此基础上 满足条件的回文串数量还要加上 dp[i+1][j-1]+1 .

因为 对于 i<t<=k<j  如果 str[t]和str[k] 组成回文串 那么  '字符M'+str[t] 和 str[k]+'字符M'又是一个新的回文串，加上本身自己又是个回文串，所以要

加上 dp[i+1][j-1]+1

 

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#define N 1005
using namespace std;
 
int dp[][];
char str[];
int main()
{
    int tcase;
    int k = ;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%s",str+);
        int len = strlen(str+);
        for(int i=;i<=len;i++) dp[i][i]=;
        for(int l=;l<=len;l++){
            for(int i=;i<=len-l+;i++){
                int j=i+l-;
                dp[i][j] = (dp[i+][j]+dp[i][j-]-dp[i+][j-]+)%;///子问题推出父问题(减掉重复子问题)
                                                                            ///有减号要加mod防止出现负数
                if(str[i]==str[j]){
                    dp[i][j] = (dp[i][j]+dp[i+][j-]+)%;///如果str[i]str[j]相等,那么会多出dp[i+1][j-1]+1个回文串
                }
            }
        }
        printf("Case %d: %d\n",k++,dp[][len]);
    }
    return ;
}