【POJ 3461】 Oulipo(KMP)

Oulipo

Time Limit: 1000MS Memory Limit: 65536K

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

题目大意：输入T组数据。每组数据两个字符串，你将第一个字符串在第二个字符串中出现的次数输出。如AZA在AZAZA中有2次，在AZAZAZA中有三次。

分析：典型的KMP问题。单模式匹配。
不懂KMP算法的请参考：
http://www.ruanyifeng.com/blog/2013/05/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm.html

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int maxn = 1e4+5;int next[maxn];char str1[maxn];char str2[1000005];/*/////简单但是会超时 void MakeNext( char *p ){    int i,k;//Q记录字符串下标 k是最大前后相同缀长度    int m = strlen(p);//字符串长度    next[0] = -1;//第一个字符串的前后缀长度为0    for( i=1 , k=0 ; i<m ; i++ ){        while( k>0 && p[i] != p[k] ){//求出p[0]-->p[i]的最大前后缀长度             k = next[k-1];        }         if( p[i] == p[k] ){            k++;        }        next[i] = k;    } } */void MakeNext( char *str ){    int n = strlen(str);    int i = 0, j = -1;    next[0] = -1;    while( i < n ){        if( j == -1 || str[i] == str[j] )            next[++i] = ++j;        else            j = next[j];    }}int kmp( char *p,char *str ){    int n = strlen(p);    int m = strlen(str);    int i = 0;    int j = 0;    int cnt = 0;    while( i<m && j<n ){        if( str[i] == p[j] || j == -1 ){            i++;            j++;        }        else{            j = next[j];        }        if( j == n ){            cnt++;            j = next[j];        }    }    return cnt;}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%s",str1);        scanf("%s",str2);        MakeNext( str1 );        int ans = kmp( str1,str2 );        printf("%d\n",ans);    }    return 0;}