HDOJ 5763 Another Meaning(kmp+dp)

Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 826    Accepted Submission(s): 383

Problem Description

As is known to all, in many cases, a word has two meanings. Such as “hehe”, which not only means “hehe”, but also means “excuse me”. 
Today, ?? is chating with MeiZi online, MeiZi sends a sentence A to ??. ?? is so smart that he knows the word B in the sentence has two meanings. He wants to know how many kinds of meanings MeiZi can express.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each test case contains two strings A and B, A means the sentence MeiZi sends to ??, B means the word B which has two menaings. string only contains lowercase letters.

Limits
T <= 30
|A| <= 100000
|B| <= |A|

 

Output

For each test case, output one line containing “Case #x: y” (without quotes) , where x is the test case number (starting from 1) and y is the number of the different meaning of this sentence may be. Since this number may be quite large, you should output the answer modulo 1000000007.

 

Sample Input

4 hehehe hehe woquxizaolehehe woquxizaole hehehehe hehe owoadiuhzgneninougur iehiehieh

 

Sample Output

Case #1: 3 Case #2: 2 Case #3: 5 Case #4: 1

Hint

In the first case, “ hehehe” can have 3 meaings: “*he”, “he*”, “hehehe”. In the third case, “hehehehe” can have 5 meaings: “*hehe”, “he*he”, “hehe*”, “**”, “hehehehe”.

 

题意：给出一个主串和子串，子串有两种含义（子串在主串中可以变成“*”），问主串一共有多种形式？

题解：令dp[i]表示到i结尾的字符串可以表示的不同含义数，那么考虑两种转移:

末尾没有出现替换含义的 ： dp[i]=dp[i-1]

末尾出现了替换含义的： dp[i]=dp[i]+dp[i-m] (m表示子串的长度)

对位第i位为结尾有没有出现替换含义（即与子串匹配），我们需要快速处理，这里使用KMP算法预处理，标记子串在主串中被完全匹配到的位置。例如对于 hehehehe    与hehe   我们标记下 4 6 8这三个位置（为了方便统计，防止出现初始位置为-1的情况，我们将下标从1开始）

代码如下：

#include
#include
#include
using namespace std;
#define maxn 100010
#define mod 1000000007
char str[maxn],s[maxn];
int f[maxn],dp[maxn],mark[maxn];
int lena,lenb,ans;

void get_next()
{
	int i=0,j=-1;
	f[0]=-1;
	while(i