HDU 5763 dp+kmp

最新推荐文章于 2018-09-06 21:31:46 发布

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本文介绍了一道关于字符串匹配的问题，通过KMP算法预处理并结合动态规划求解字符串的不同含义数量。具体包括如何利用KMP算法进行字符串匹配，以及如何通过DP算法计算可能的意义数目。

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Another Meaning

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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：给你t个测试样例每个测试样例两个字符串上面那个是原字符串下面是这些字符可以用第二个意思替换问这句话有几种意思

思路：先用kmp处理好有多少重叠匹配的串，弄为一个01串，定义dp[i]为当前这里有多少种意思，所以当前可以匹配的话，则dp[i]=dp[i-1]+1，如果当前的串不影响前面的串，那么就可以同时匹配，所以dp[i]=dp[i-1]+dp[i-m]+1，如果当前的串不能匹配，则dp[i]=dp[i-1]，最后输出要输出dp[n-1]+1，因为当所有能匹配的地方都是第一个意思，所以要+1.

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef __int64 ll;
char s1[100005],s2[100005],nex[100005];
ll vis[100005],dp[100005];
int main(){
    ll t,i,j,cas=1;
    scanf("%I64d",&t);
    while(t--){
        scanf("%s%s",s1,s2);// s2 是匹配串 
        memset(nex,-1,sizeof(nex)); //初始化为-1 
        memset(vis,0,sizeof(vis));
        memset(dp,0,sizeof(dp));
        ll n=strlen(s1); 
        ll m=strlen(s2);
        j=-1;
        for(i=1;i<m;i++){
            while(j>=0&&s2[j+1]!=s2[i])j=nex[j];
            if(s2[j+1]==s2[i])j++;
            nex[i]=j;
        }    
        j=-1;
        for(i=-1;i<n-1;i++){
            while(j>=0&&s2[j+1]!=s1[i+1])j=nex[j];
            if(s2[j+1]==s1[i+1])j++;
            if(j==m-1){
                //j=next[j]; //重叠匹配 
                //j=-1; //从头匹配 
                j=nex[j];
                vis[i+1]=1;
            }
        }
        ll num=0;
        for(i=0;i<m;i++){
        	dp[i]=vis[i];
        }
        for(;i<n;i++){
        	if(vis[i]){
	        	dp[i]=(dp[i-1]+dp[i-m]+1)%1000000007;
	        }
	        else dp[i]=dp[i-1];
        }
        printf("Case #%I64d: %I64d\n",cas++,(dp[n-1]+1)%1000000007);
    }
    return 0;
}