HDU 1686 Oulipo (KMP查找可交叉區間模式串)

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

唯一的修改就是在返回位置的地方修改爲j=f[j]

#include <string.h>
#include <iostream>
#include <cstdio>
#define maxn 1000010
int next[maxn];
void getNext(char* P, int* f)
{
    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;
    }
}

int kmp(char* T, char*P, int*f)
{
    int n = strlen(T), m = strlen(P);
    getNext(P, f);
    int ans=0;
    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)
        {
            ans++;//i-m+1
            j=f[j];
        }
    }
    return ans;
}
int main()
{
    char s1[maxn];
    char s2[maxn];
    int a;
    int n;
    int m;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",s2,s1);
        int a=kmp(s1,s2,next);
        printf("%d\n",a);
    }
    return 0;
}