HDU5763 Another Meaning动态规划

题目链接：HDU5763

Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 843    Accepted Submission(s): 391

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：有一个字符串t有二义性，问字符串s有多少种解释方法（具体的见hint吧）。

题目分析：动态规划，设dp[i]为以i结尾的字符串有多少种语义，状态转移方程dp[i]=dp[i-1]+dp[i-lt];dp[i-1]为当以i结尾的字串不翻译为第二个意思时，dp[i-lt]就是翻译为第二个意思了。可以先用kmp把所有可能产生二义的位置hash出来，注意i-lt可能会出现-1的情况，把它替换成0就行了。

//
//  main.cpp
//  Another Meaning
//
//  Created by teddywang on 2016/7/28.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[100010],t[100010];
int nexts[100010];
int ls,lt;
int p[100010];
int dp[100010];
int mod=1000000007,ans,num;
void getnext()
{
    int i=0,j=-1;
    nexts[0]=-1;
    while(i<lt&&j<lt)
    {
        if(j==-1||t[i]==t[j])
        {
            nexts[++i]=++j;
        }
        else j=nexts[j];
    }
}

void kmp()
{
    int i=0,j=0;
    while(i<ls&&j<lt)
    {
        if(j==-1||s[i]==t[j])
        {
            ++i;++j;
            if(j==lt)
            {
                p[i-1]=1;
                j=nexts[j];
            }
        }
        else j=nexts[j];
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    for(int k=1;k<=T;k++)
    {
        memset(dp,0,sizeof(dp));
        memset(p,0,sizeof(p));
        scanf("%s%s",s,t);
        ls=strlen(s),lt=strlen(t);
        getnext();
        kmp();
        if(p[0]==1) dp[0]=2;
        else
        dp[0]=1;
        for(int i=1;i<ls;i++)
        {
            dp[i]=(dp[i-1]+p[i]*dp[max(i-lt,0)])%mod;
        }
        printf("Case #%d: %d\n",k,dp[ls-1]);
    }
}