POJ3461[Oulipo]KMP裸题

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

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solution: KMP

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;#define N 1000007#define M 10007char t[M], s[N];int nxt[M];int lent, lens, tot;void get_next(){    int i=0, j=-1;    nxt[0]=-1;    while( i<lent ){        if ( j==-1||t[i]==t[j] ) i++, j++, nxt[i]=j;        else j=nxt[j];    }}void Kmp(){    int i=0, j=0;    while( i<lens ){        if( j==-1||s[i]==t[j] ){            i++, j++;            if( j==lent ) tot++, j=nxt[j];        }else j=nxt[j];    }}int main(){    int T;    scanf( "%d", &T);    while( T-- ){        tot=0;        scanf( "%s%s", t, s);        lent=strlen(t);        lens=strlen(s);        get_next(); Kmp();        printf( "%d\n", tot);    }    return 0;}