[poj 3461]Oulipo[kmp]

Oulipo
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 34693 Accepted: 14012

Description

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

Source
BAPC 2006 Qualification

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>using namespace std;const int maxn = 1000010;int T,cnt;int next[maxn];char t[maxn],p[maxn];void getnext(){    int lp = strlen(p);    next[0]= -1;next[1]=0;    for(int i=1;i<lp;i++)    {        int j = next[i];        while(j&&p[i]!=p[j])j=next[j];        next[i+1]=p[i]==p[j]?j+1:0;    }}void kmp(){    int lt = strlen(t),lp = strlen(p);    getnext();    int j = 0;    for(int i= 0;i<lt;i++)    {        while(j&&p[j]!=t[i])j=next[j];        if(p[j]==t[i])j++;        if(j==lp){            cnt++;            j = next[j];        }    }}int main(){    scanf("%d",&T);    while(T--)    {        memset(next,0,sizeof(next));        scanf("%s",p);        scanf("%s",t);        cnt = 0;        kmp();        printf("%d\n",cnt);    }    return 0;}

    #include<iostream>      #include<cstring>      #include<cstdio>      #include<string>      #include<algorithm>      using namespace std;      #define N 100010      char s[N * 10], p[N];      int nextval[N];      int lens, lenp;      void getnext()      {          int i = 0, j = -1;          nextval[0] = -1;          while(i != lenp)          {              if(j == -1 || s[i] == s[j])                  nextval[++i] = ++j;              else                  j = nextval[j];          }      }      int KMP()      {          int i = 0, j = 0, count = 0;          while(i != lens && j != lenp)          {              if(s[i] == p[j] || j == -1)                  ++i, ++j;              else                  j = nextval[j];              if(j == lenp)              {                  count++;                  j = nextval[j];              }          }          return count;      }      int main()      {          int ncase;          int len;          int ans;          scanf("%d", &ncase);          while(ncase--)          {              scanf("%s%s", p, s);              lens = strlen(s);              lenp = strlen(p);              getnext();              ans = KMP();              printf("%d\n", ans);          }          return 0;      }