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 = <Sx1, Sx2, ..., Sxk> and Y = <Sy1, Sy2, ..., Syk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if Sxi = Syi. Also two subsequences with different length should be considered different.
(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 = <Sx1, Sx2, ..., Sxk> and Y = <Sy1, Sy2, ..., Syk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if Sxi = Syi. 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
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[1005][1005]; char str[1005]; int main() { int tcase; int k = 1; scanf("%d",&tcase); while(tcase--){ scanf("%s",str+1); int len = strlen(str+1); for(int i=1;i<=len;i++) dp[i][i]=1; for(int l=2;l<=len;l++){ for(int i=1;i<=len-l+1;i++){ int j=i+l-1; dp[i][j] = (dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+10007)%10007;///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+1][j-1]+1)%10007;///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 ",k++,dp[1][len]); } return 0; }