HDU 5763：Another Meaning（字符串匹配）

http://acm.hdu.edu.cn/showproblem.php?pid=5763

Another Meaning

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：每个例子的第二个串有两个意思，然后求第一个串可以表达多少种意思。

思路：做的时候DP方程推不出来，太渣了。dp[i]表示以i结尾长度的串有多少种意思，如果不替换的话，dp[i] = dp[i-1]，如果当前匹配的话可以替换，要加上前面的dp[i-m]，赛后补题的时候发现string类有个substr的函数很厉害。

 

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define N 100005
#define MOD 1000000007
/*
对于这个问题，显然可以进行DP：
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:

末尾不替换含义：dp[i - 1]

末尾替换含义：dp[i - |B|]  (A.substr(i - |B| + 1,|B|) = B)

那么对于末尾替换含义的转移，需要快速判断BB能不能和当前位置的后缀匹配，kmp或者hash判断即可。

复杂度:O(N)
*/
long long dp[N];
string s, str;

int main()
{
    int t;
    cin >> t;
    for(int cas = 1; cas <= t; cas++) {
        memset(dp, 0, sizeof(dp));
        cin >> s >> str;
        int n = s.size();
        int m = str.size();
        for(int i = 0; i < m; i++)
            dp[i] = 1;
        for(int i = m; i <= n; i++) {
            dp[i] = dp[i-1];
            string ss = s.substr(i - m, m);
            if(ss == str)
                dp[i] += dp[i-m] % MOD;
            dp[i] %= MOD;
        }
        printf("Case #%d: ", cas);
        cout << dp[n] << endl;
    }
    return 0;
}