2016多校联赛4A Another Meaning（hdu 5763）

Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 266    Accepted Submission(s): 114

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

题意：就是给你两个字符串，第一个字符串与第二个字符串相同的地方可以改变意思，也可以不改变意思,问最后第一个字符串有多少种理解方案。

思路：这道题要用dp。dp[i]表示为以i结尾时有多少第一个字符串有多少种含义。然后我们应该怎样dp呢？分两种情况。如果当前位置是能和第二个字符串完全匹配的，也就是说这个地方是可以改变意思的。那么我们有两种状态，一是不改变这里的意思，二是改变这里的意思。不改变意思的方案有dp[i-1]种，改变意思的方案是有dp[i-strlen(s2)]种情况，这个考虑到前面如果改变这里就不能改变的问题就很好理解。显然第二种情况就是不能改变意思，不能改变意思就等于dp[i-1],那么如何判断这里能不能改变意思呢？我们用到了kmp算法标记位置。剩下的也没什么了。。。下面给代码。

#include<iostream>
#include<stack>
#include<cstring>
#include<map>
#include<string>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
#define maxn 100005
#define MOD 1000000007
char s1[maxn], s2[maxn];
int next1[maxn];
int dp[maxn], vis[maxn],tcase;
void makeNext()
{
	int q, k;
	int m = strlen(s2);
	next1[0] = 0;
	for (q = 1, k = 0; q < m; ++q)
	{
		while (k > 0 && s2[q] != s2[k])
			k = next1[k - 1];         
		if (s2[q] == s2[k])
		{
			k++;
		}
		next1[q] = k;
	}
}
void kmp()
{
    int n, m;
    int i, q;
    n = strlen(s1);
    m = strlen(s2);
    for (i = 0, q = 0; i < n; ++i)
    {
        while (q > 0 && s2[q] != s1[i])
            q = next1[q - 1];
        if (s2[q] == s1[i])
        {
            q++;
        }
        if (q == m)
        {
			vis[i] = 1;
        }
    }
	if (vis[m - 1]){
		dp[m - 1] = 2;
	}
	else
		dp[m - 1] = 1;
	for (int i = m ; i < n; i++){
		if (vis[i]){
			dp[i] += max(dp[i - m], 1);
			dp[i] %= MOD;
			dp[i] += max(dp[i - 1], 1);
			dp[i] %= MOD;
		}
		else{
			dp[i] += dp[i - 1];
			dp[i] %= MOD;
		}
	}
	printf("Case #%d: %d\n", tcase, dp[n - 1]);
}
int main(){
	int t;
	scanf("%d", &t);
	for (tcase = 1; tcase <= t; tcase++){
		scanf("%s%s", s1, s2);
		makeNext();
		memset(dp, 0, sizeof(dp));
		memset(vis, 0, sizeof(vis));
		kmp();
	}
}