解题报告：HDU 5763

最新推荐文章于 2017-01-15 19:42:52 发布

HyJoker

最新推荐文章于 2017-01-15 19:42:52 发布

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本文链接：https://blog.youkuaiyun.com/HyJoker/article/details/52066209

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本文介绍了一种使用动态规划解决含有双义词的句子可能存在的不同含义数量的问题。通过调整DP的状态定义，实现了一个高效的算法。

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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


    
    
     
     
      
      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”.

这道题很明显是用DP做，但是最主要的是DP的方式……

最开始的想法是dp[i]存放从0开始到i为止的种类数,然后分为本意和另外一个意思来分别讨论，结果发现这样做不出来……

后来看过题解才明白只要转换下思路就可以了，dp[i]应该表示位置i到末尾的种类数。

首先dp[i]=dp[i-1],即每个当前位置都是由前一个位置得出来的，这是增加的基础。

其次如果i-len2到i这段和b匹配上了，那如果他使用了另一个含义，则i-len2到i的都不能改变了，所以应该是加上i-len2时候的种类数。

所以dp[i]+=dp[i-len2]

代码其实很简单，如下

#include<iostream>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;
string a,b;
long long dp[100010];
int mod=1e9+7;
int main()  
{  
    //freopen("in.txt","r",stdin);  
    //freopen("out.txt","w",stdout);  
    int t;  
    cin>>t;
    for(int Case=1;Case<=t;Case++)  
    {  
		memset(dp,-1,sizeof(dp));
		cin>>a>>b;
		dp[0]=1;
		for(int i=1;i<=a.length();i++){

			dp[i]=dp[i-1];
			if(i>=b.length()&&a.substr(i-b.length(),b.length())==b)
				dp[i]=(dp[i-b.length()]+dp[i])%mod;

		}
		printf("Case #%d: %d\n",Case,dp[a.length()]);
		
    }  
    return 0;  
}

所以说学好DP还是很重要的……