HDU 4632 Palindrome subsequence 记忆化搜索

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蒟蒻的ACMer

最新推荐文章于 2024-02-03 13:30:00 发布

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分类专栏： D P一般看规律文章标签： c++ HDU DP

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本文介绍了一种算法，用于解决给定字符串中不同回文子串的数量问题。通过动态规划方法，文章提供了一个高效的解决方案，并附带了完整的代码实现。

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Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2735 Accepted Submission(s): 1105

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

给出一个串问区间有多少个回文子串

dp[i][j]表示第i个到第j个有dp[i][j]个子串

当str[i]==str[j]时候可以知道dp[i][j]=(dp(i,j-1)+dp(i+1,j)+)%mod

否则dp[i][j]=(dp[i][j-1]+dp[i+1][j]-dp[x+1][y+1]+mod)%mod可能出现负值

ACcode:

#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define maxn 1001
#define mod 10007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI  acos(-1.0)
#define E  exp(1)
using namespace std;
int loop,dp[maxn][maxn],cnt=1,len;
char str[1010];
int DP(int x,int y){
    if(x>y)return 0;
    if(x==y)return 1;
    if(dp[x][y])return dp[x][y];
    if(str[x-1]==str[y-1])
       return  dp[x][y]=(DP(x,y-1)+DP(x+1,y)+1)%mod;
    return dp[x][y]=(DP(x,y-1)+DP(x+1,y)-DP(x+1,y-1)+mod)%mod;
}
int main(){
    rd(loop);
    while(loop--){
        rds(str);MT(dp,0);
        len=strlen(str);
        printf("Case %d: %d\n",cnt++,DP(1,len));
    }
    return 0;
}
/*
4
a
aaaaa
goodafternooneveryone
welcometoooxxourproblems
*/