HDU 5763 Another Meaning KMP+DP

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 <= 30 > |A| <= 100000 > |B| <= |A| #### 输出

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;
}