KMP算法模板

蒟蒻学识浅陋,欢迎各位大牛指正

KMP从入门到放弃

请观左神为什么想要杀人%%%%njb7
着重听1h12m20s
$KMP$分为两个部分,一部分为两个字符串间的比较,另一部分为自己与自己的比较.
简单的划分为下面两个图,详细理解请见左神不稳定情绪讲解.


不过我$jiao$的在$1:21:04$时,将例子换为$”ababcababak”$更好理解

代码

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stdio.h>
#include<stdlib.h>
#ifndef NULL
#define NULL 0
#endif
using namespace std;

char t[1000100],s[1000100];
int len1, len2,n[1000100];
void KMP(char *s,char *t)
{
    for (int i = 0, j = -1; i < len1; i++) {
        while (j != -1 && t[j + 1] != s[i])
            j = n[j];
        if (t[j + 1] == s[i])
            j++;
        if (j == len2 - 1) {
            cout << i - len2 + 2 << endl;
            j = n[j];
        }
    }
}
void getnext(char *t)
{
    n[0] = -1;
    for (int i = 1, j = -1; i < len2; i++) {
        while (j != -1 && t[i] != t[j + 1])
            j = n[j];
        if (t[i] == t[j + 1])
            j++;
        n[i] = j;
    }
}
int main()
{
    cin >> s >> t;
    len1 = strlen(s);
    len2 = strlen(t);
    getnext(t);
    KMP(s, t);
    for (int i = 0; i < len2; i++)
        cout << n[i]+1 << ' ';
    return 0;
}

例题

G.Oulipo

HDU 1686
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 $500000$ 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 $1 ≤ |W| ≤ 10000$ (here $|W|$ denotes the length of the string $W$).
One line with the text T, a string over ${‘A’, ‘B’, ‘C’, …, ‘Z’}$, with $|W| ≤ |T| ≤ 1000000$.

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

解析
$kmp$模板,读入不能用$cin$,否则超时

#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
#include<stdio.h>
#include<stdlib.h>
#ifndef NULL
#define NULL 0
#endif
using namespace std;

typedef long long ll;
char t[1000100], s[1000100];
int len1, len2, n[1000100];
ll read()
{
    ll f = 1,x = 0;
    char s = getchar();
    while (s<'0' || s>'9') {
        if (s == '-')
            f = -1;
        s = getchar();
    }
    while (s >= '0'&&s <= '9') {
        x = x * 10 + s - '0';
        s = getchar();
    }
    x *= f;
    return x;
}
int KMP(char *s, char *t)
{
    int ans = 0;
    for (int i = 0, j = -1; i < len1; i++) {
        while (j != -1 && t[j + 1] != s[i])
            j = n[j];
        if (t[j + 1] == s[i])
            j++;
        if (j == len2 - 1) {
            ans++;
            j = n[j];
        }
    }
    return ans;
}
void getnext(char *t)
{
    n[0] = -1;
    for (int i = 1, j = -1; i < len2; i++) {
        while (j != -1 && t[i] != t[j + 1])
            j = n[j];
        if (t[i] == t[j + 1])
            j++;
        n[i] = j;
    }
}
int main()
{
    int m;
    m = read();
    while (m--) {
        memset(n, 0, sizeof(n));
        scanf("%s %s", &t, &s);
        len1 = strlen(s);
        len2 = strlen(t);
        getnext(t);
        cout<<KMP(s, t)<<endl;
    }
    return 0;
}