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 = <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
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;
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #define MAX 1010
5 #define MOD 10007
6 using namespace std;
7 int dp[MAX][MAX];
8 char str[MAX];
9 int main(){
10 int T, cnt = 0;
11 //freopen("in.c", "r", stdin);
12 scanf("%d", &T);
13 while(T--){
14 memset(str, 0, sizeof(str));
15 memset(dp, 0, sizeof(dp));
16 scanf("%s", str);
17 int len = strlen(str);
18 for(int i = 0; i< len;i ++) dp[i][i] = 1;
19 for(int i = 0;i < len;i ++){
20 for(int j = i-1;j >= 0;j --){
21 dp[j][i] = (dp[j][i-1] + dp[j+1][i] - dp[j+1][i-1] + MOD)%MOD;
22 if(str[i] == str[j]) dp[j][i] = (dp[j][i] + dp[j+1][i-1] + 1 + MOD)%MOD;
23 }
24 }
25 printf("Case %d: %d\n", ++cnt, dp[0][len-1]);
26 }
27 return 0;
28 }
HDOJ -- 4632 区间DP,布布扣,bubuko.com
HDOJ -- 4632 区间DP
原文：http://www.cnblogs.com/anhuizhiye/p/3618702.html
