HDU 5763 Another Meaning KMP+DP

时间：2024-09-25 21:36:14

题目链接：

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

Another Meaning

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)
#### 问题描述
> 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.
#### 输入
> 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 |A| |B|

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.

样例

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

题解

dp

用kmp处理出匹配成功的结尾的字母位置，然后就是考虑每个位置选和不选的两种情况，转移下就可以了。

代码

#include<map>

#include<queue>

#include<vector>

#include<cstdio>

#include<cstring>

#include<string>

#include<iostream>

#include<algorithm>

#define X first

#define Y second

#define mkp make_pair

#define lson (o<<1)

#define rson ((o<<1)|1)

#define M (l+(r-l)/2)

#define bug(a) cout<<#a<<" = "<<a<<endl 

using namespace std;

typedef __int64 LL;

const int maxn=1e5+10;

const int INF=0x3f3f3f3f;

const LL INFL=0x3f3f3f3f3f3f3f3fLL;

const int mod=1e9+7;

char s1[maxn],s2[maxn];

int vis[maxn];

vector<int> pos;

LL dp[maxn];

int f[maxn];

void getFail(char *P){

	int m=strlen(P);

	f[0]=0; f[1]=0;

	for(int i=1;i<m;i++){

		int j=f[i];

		while(j&&P[i]!=P[j]) j=f[j];

		f[i+1]=P[i]==P[j]?j+1:0;

	}

}

void find(char* T,char *P){

	int n=strlen(T),m=strlen(P);

	getFail(P);

	int j=0;

	for(int i=0;i<n;i++){

		while(j&&P[j]!=T[i]) j=f[j];

		if(P[j]==T[i]) j++;

		if(j==m){

			pos.push_back(i+1);

			vis[i+1]=1;

		}

	}

}

void init(){

	pos.clear();

	memset(vis,0,sizeof(vis));

}

int main() {

	int tc,kase=0;

	scanf("%d",&tc);

	while(tc--){

		init();

		scanf("%s%s",s1,s2);

		find(s1,s2);

		memset(dp,0,sizeof(dp));

		dp[0]=1;

		int n=strlen(s1),m=strlen(s2);

		for(int i=1;i<=n;i++){

			//第i位不选

			dp[i]=dp[i-1];

			//第i位选

			if(vis[i]) dp[i]+=dp[i-m];

			dp[i]%=mod;

		}

		printf("Case #%d: %I64d\n",++kase,dp[n]);

	}

	return 0;

}

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