2016多校训练Contest4: 1001 Another Meaning hdu5763

最新推荐文章于 2019-06-14 18:12:55 发布

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#KMP #DP

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本文介绍了一种利用KMP算法进行字符串匹配，并结合动态规划（DP）来解决特定字符串问题的方法。通过实例演示了如何计算一个句子中带有双义词语的不同含义的数量。

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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

HintIn 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”.

先KMP求出匹配的位置

然后直接dp转移

f[i]=f[i-1]

if(v[i-B.size()+1])

f[i]+=f[i-B.size(]]

v[]表示该位是否为匹配位

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int mod=1000000007;
int loc[1000001];
int ne[1000001];
int ff[1000001];
bool v[100001];
string test,pattern;
int m,n;
inline void kmp()
{
     int i,j;
     for(i=1;i<n;i++)
     {
          j=i;
          while(j>0)
          {
               j=ne[j];
               if(pattern[j]==/*test[i]*/pattern[i])
               {
                    ne[i+1]=j+1;
                    break;
               }
          }
     }
     for(i=0,j=0;i<m;i++)
     {
          if(j<n&&test[i]==pattern[j])
               j++;
          else
          {
               while(j>0)
               {
                    j=ne[j];
                    if(test[i]==pattern[j])
                    {
                         j++;
                         break;
                    }
               }
          }
          if(j==n)
          {
               loc[0]++;
               loc[loc[0]]=(i-n+1)+1;
          }
     }
}
int main()
{
	int k=0;
	int T;
	scanf("%d",&T);
	while(T>0)
	{
		T--;
		k++;
		cin>>test>>pattern;
		m=test.size();
		n=pattern.size();
		memset(ne,0,sizeof(ne));
		memset(loc,0,sizeof(loc));
		memset(ff,0,sizeof(ff));
		kmp();
		int i;
		printf("Case #%d: ",k);
		if(loc[0]==0)
			printf("1\n");
		else
		{
			memset(v,false,sizeof(v));
			for(i=1;i<=loc[0];i++)
				v[loc[i]]=true;
			ff[0]=1;
			for(i=1;i<=m;i++)
			{
				ff[i]=ff[i-1]%mod;
				if(i-n>=0)
					if(v[i-n+1])
						ff[i]=(ff[i]+ff[i-n])%mod;
			}
			printf("%d\n",ff[m]);
		}
	}
	return 0;
}