KMP---POJ 3461Oulipo

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

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

Sample Output

130

题目大意:找重复的子串

#include<iostream>#include<string>#include<vector>using namespace std;int kmp_find(const string& target,const string& pattern){    const int target_length = target.size();    const int pattern_length = pattern.size();    int * overlay_value = new int[pattern_length];    overlay_value[0] = -1;    int index = 0;//得到匹配值    for(int i=1;i<pattern_length;++i)    {        index = overlay_value[i-1];        while(index>=0 && pattern[index+1]!=pattern[i])        {            index  = overlay_value[index];        }        if(pattern[index+1]==pattern[i])        {            overlay_value[i] = index +1;        }        else        {            overlay_value[i] = -1;        }    }    //match algorithm start    int pattern_index = 0;//用来小串的移动    int target_index = 0;//用来大串的移动int sum=0;//统计一个几个    while(target_index<target_length)    {if(target[target_index]==pattern[pattern_index])        {//如果匹配就继续前移            ++target_index;            ++pattern_index;        }else{//如果不匹配//这里注意下pattern_index=0的情况//如果为0，然后上一步又不匹配，那么直接让++target_index;if(pattern_index==0){++target_index;}//否则让pattern_index实现跳else pattern_index = overlay_value[pattern_index-1]+1;}if(pattern_index==pattern_length){sum++;pattern_index = overlay_value[pattern_index-1]+1;}//注意这一步    }delete [] overlay_value;return sum;    }int main(){int t;cin>>t;string source,pattern;while(t--){    cin>>pattern>>source;    cout<<kmp_find(source,pattern)<<endl;    }return 0;}