B站的KMP演算法講解:

視訊1

視訊2

視訊中的程式碼

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
using namespace std;

void prefix_table(char pattern[], int prefix[], int n) {
	prefix[0] = 0;
	int len = 0;
	int i = 1;
	while(i < n) {
		if(pattern[i] == pattern[len]) {
			len++;
			prefix[i] = len;
			i++;
		}
		else {
			if(len > 0) {
				len = prefix[len-1];
			}
			else {
				prefix[i] = len;
				i++;
			}
		}
	}

}
void move_prefix_table(int prefix[], int n) {
	for(int i = n-1; i > 0; i--) {
		prefix[i] = prefix[i-1];
	}
	prefix[0] = -1;
}
void kmp(char text[], char pattern[]) {
	int n = strlen(pattern);
	int m = strlen(text);
//	int* prefix = malloc(sizeof(int) * n); //C語言可以用,c++用不了
	int* prefix = (int*)malloc(sizeof(int)*n); //int *prefix = new int;
	prefix_table(pattern, prefix, n);
	move_prefix_table(prefix, n);

	//text[i]  len(text) = m
	//pattern[j]   len(pattern)=n

	int i = 0;
	int j = 0;
	while(i < m) {
		if(j == n - 1 && text[i] == pattern[j]) {
		 	printf("Find pattern at %d\n", i - j);
			j = prefix[j];
		}
		if(text[i] == pattern[j]) {
			i++;
			j++;
		}
		else {
			j = prefix[j];
			if(j == -1) {
				i++;
				j++;
			}
		}
	}
}

int main() {
	char pattern[] = "ABABCABAA";
	char text[] = "ABABABCABAABABABAB";
	kmp(text, pattern);

	return 0;
}
 

暴力匹配法:

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
using namespace  std;
int main() {
    string m;
    string s;
    cout<<"請輸入主串:\n";
    cin>>m;
    cout<<"請輸入子串:\n";
    cin>>s;
    int cnt = 0;
    int flg;
    for (int j = 0; j <= (m.size() - s.size()); j++) {
        for (int i = 0; i < s.size(); i++) {
            flg = j;
            if (s[i] == m[j++]) {
                cnt++;
            }
            else {
                cnt = 0;
                i = 0;
                j = flg;
                break;
            }
        }
    }
    if (cnt == s.size()) {
        cout<<"匹配\n";
    }
    else {
        cout<<"不匹配\n";
    }
    return 0;
}
 

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+13;
const int M=1e4+13;
int n,m;//n為長串長度,m為短串長度
int a[N];//長串
int b[M];//短串
int prefix[M];

void prefix_table(){
	int i=0;
	int j=1;
    prefix[0]=0;
	while(j<m){
		if(b[j]==b[i]){
			prefix[j]=++i;
			j++;
		}
		else if(!i){
            prefix[j]=0;//不知道為什麼不加這一句,打印出來也是0
			j++;
		}
		else{
			i=prefix[i-1];
		}
	}
//	for(int j=0;j<m;j++) //prefix[]數組裡到底裝的是啥
//		cout<<prefix[j]<<" ";
//	cout<<endl;
}
int kmp(){
	int i=0;
	int j=0;
	while(i<n&&j<m){
		if(a[i]==b[j]){
			i++;
			j++;
		}
		else if(!j){
			i++;
		}
		else{
			j=prefix[j-1];
		}
	}
	if(j==m)
		return i-m+1;
	else
		return -1;
}
int main(){
	int t;
	cin>>t;
	while(t--){
		cin>>n>>m;
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(int i=0;i<m;i++)
			scanf("%d",&b[i]);
		prefix_table();
		cout<<kmp()<<endl;
	}
	return 0;
}

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

題意:子串在主串中出現的次數

提交超時

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

string W;
string T;//T是主串
int prefix[10005];

void prefix_table() {
    int i = 0;
    int j = 1;
    prefix[0] = 0;
    while (j < W.size()) {
        if (W[j] == W[i]) {
            prefix[j] = ++i;
            j++;
        }
        else if (!i) {
            j++;
        }
        else {
            i = prefix[i-1];
        }
    }
    return ;
}

int kmp() {
    int i = 0;
    int j = 0;
    int cnt = 0;
    while (i < T.size()) {
        if (T[i] == W[j]) {
            i++;
            j++;
        }
        else if (!j) {
            i++;
        }
        else {
            j = prefix[j-1];
        }
        if (j == W.size()) {
            cnt++;
            j = prefix[j-1];
        }
    }
    return cnt;
}

int main() {
    int n;
    cin>>n;
    while (n--) {
        //先輸入子串,再輸入主串
        cin>>W;
        cin>>T;//主串
        prefix_table();
        cout<<kmp()<<endl;
    }
    return 0;
}

下面這個也超時,納悶

#include <iostream>
#include <string>
#include <string.h>
using namespace std;
string a,b;//a是主串
int nxt[1005];

void getnxt() {
    memset(nxt, 0, sizeof(nxt));
    nxt[0] = -1;
    int k = -1;
    int i = 0;
    int len = b.size();
    while (i < len) {
        if (k == -1 || b[i] == b[k]) {
            i++;
            k++;
            nxt[i] = k;
        }
        else
            k = nxt[k];
    }
}

int kmp() {
    int cnt = 0;
    int i = 0;
    int j = 0;
    int lena = a.size();
    int lenb = b.size();
    while (i < lena) {
        if (j == -1 || a[i] == b[j]) {
            i++;
            j++;
        }
        else
            j = nxt[j];
        if (j == lenb) {
            cnt++;
            j = 0;
            i = i - lenb +1;
        }
    }
    return cnt;
}

int main() {
    int n;
    cin>>n;
    while (n--) {
        cin>>b;
        cin>>a;//主串
        int len = b.size();
        getnxt();
        cout<<kmp()<<endl;
    }
    return 0;
}