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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:
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
思路：
KMP算法的实现就能AC，朴素匹配算法会超时用不得。
我的代码：
#include <iostream> #include <cstdio> #include <cstring> using namespace std; char a[10010],b[1000010]; int ne[10010]; void fun(char *p,int L)//找出子串的相同的前缀与后缀，用ne数组储存起来 { int i,j; i=0;j=-1;//大话数据结构的是从ne【1】开始的，但是我不会忽略ne【0】输入一个字符串 ne[0]=-1; while (i<L) { if (j==-1||p[i]==p[j]) { i++;j++; ne[i]=j; } else { j=ne[j]; } } } int main() { int n,i,j,k,lena,lenb,count; scanf("%d",&n); while (n--) { scanf("%s %s",&a,&b); lena=strlen(a); fun(a,lena); lenb=strlen(b); count=0; i=-1;j=-1; while (i<lenb)//这里与寻找ne数组很像 { if (j==-1||a[j]==b[i]) { i++; j++; } else { j=ne[j]; } if (j==lena) { count++; j=ne[j]; } } printf("%d\n",count); } return 0; }
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原文地址：http://blog.csdn.net/zsc2014030403015/article/details/45270469
