Oulipo（KMP）

Oulipo

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter ‘e’. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T’s is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {‘A’, ‘B’, ‘C’, …, ‘Z’} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

One line with the word W, a string over {‘A’, ‘B’, ‘C’, …, ‘Z’}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
One line with the text T, a string over {‘A’, ‘B’, ‘C’, …, ‘Z’}, with |W| ≤ |T| ≤ 1,000,000.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

Sample Input
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0

这道题将两个字符串合成一个，中间加一个符号，子字符串是该符号的前缀，往后求其后缀，当后缀长度达到前缀的长度，即符号后面的字符串包含该子字符串。
KMP：个人理解就是找到将字符串中遍历，求出每个字符串与首字符串的公共序列的长度

ac代码：（KMP）

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
using namespace std;

string s,t;
int n,m;
vector<int> cal_nxt(string s){
	int n=(int)s.size();
	vector<int> nxt(n);
	for(int i=1;i<n;i++){
		int j=nxt[i-1];
		while(s[i]!=s[j]&&j>0) j=nxt[j-1];//向前缩短序列，以致s[i]==s[j] 
		if(s[i]==s[j]) j++;
		nxt[i]=j;
	}
	return nxt;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0),cout.tie(0);
	int tt;
	cin>>tt;
	while(tt--){
		cin>>t>>s;
	m=t.size();
	n=s.size();
	string T=t+'#'+s;//将两个字符串相加成为一个，求#得最长前后缀
	vector<int> nxt=cal_nxt(T);//
	int flag=0;
	for(int i=m+1;i<(int)T.size();i++){
		if(nxt[i]==m){
			flag++;
		}
	} 
	cout<<flag<<endl;
	}
	
	return 0;
}

ac代码：（哈希）

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
typedef unsigned long long ull;
const ull base =2333;//进制
const ull mod = 1e9+7;//取余防止爆精度
const int N=1e6+100;
ull hashes[N],p[N];
ull gethashes(int l,int r){
	return (hashes[r]%mod-(hashes[l-1]%mod*p[r-l+1]%mod)%mod+mod)%mod;
	
} 
char s[N],t[N];                               


int main(){
	int T;
	cin>>T;
	while(T--){
		scanf("%s %s",t+1,s+1);//初始位置为1 
		int n=strlen(s+1);
		int m=strlen(t+1);
		p[0]=1;
		for(int i=1;i<=n;i++){
			hashes[i]=(hashes[i-1]*base%mod+s[i]%mod)%mod;
			p[i]=(p[i-1]%mod*base%mod)%mod;
		}
		ull ans=0;
		for(int i=1;i<=m;i++) ans=(ans*base%mod+t[i]%mod)%mod;
		int res=0;
		for(int i=m;i<=n;i++){
			if(gethashes(i-m+1,i)==ans){
				res++;
			}
		}
		printf("%d\n",res);
	} 
	
	
	return 0;
}