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 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.

InputThe 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.
OutputFor 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

http://blog.csdn.net/v_july_v/article/details/7041827（kmp的算法详解）

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

int n;
int nxt[1000005];
char p[10005],s[1000005];

void getNext(char *p){
    nxt[0]=-1;
    int len=strlen(p);
    for(int i=0;i<len;i++){
        int k=nxt[i];
        while(k!=-1&&p[i]!=p[k]) k=nxt[k];
        nxt[i+1]=k+1;
    }
}

int kmp(char *s,char *p){
    getNext(p);
    int ans=0,i=0,j=0,lens=strlen(s),lenp=strlen(p);
    while(i<lens){
        if(j==-1||s[i]==p[j]){
            i++;
            j++; 
        }
        else j=nxt[j];
        if(j==lenp){
            ans++;
            j=nxt[j];
        }
    }
    return ans;
}

int main()
{   scanf("%d",&n);
    while(n--){
        scanf("%s%s",p,s);
        printf("%d\n",kmp(s,p));
    }
}