POJ 3461&& hiho1015 Oulipo [KMP算法]

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

题意：读入一个子串P，再读入一个主串S，问S串中有多少个P；

题解：典型的KMP算法裸题，直接套板子就OK（然而我并不会）；

AC代码：

#include <iostream>#include <math.h>#include <stdio.h>#include <string.h>#include <algorithm>#define ll long long#define N 100010using namespace std ;char s[N*10] , p[N] ;int next[N*10] , lenp , lens;//*lenp是匹配，lens是主void kkk(){    int i , j ;    i = 0 ;    j = next[0] = -1;    while(i < lenp) //*主串里面找匹配    {        while(j!=-1&&s[i]!=s[j])            j = next[j];        next[++i] = ++j ;    }}int KMP(){    int i , j , ans ;    i = ans = j = 0 ;    while(i!=lens&&j!=lenp)    {        if(s[i]==p[j]||j==-1)            ++i , ++j;        else j = next[j];        if(j==lenp)        {            ans++;            j = next[j] ;        }    }    return ans ;}int main(){    int t ;    cin>>t;    while(t--)    {        memset(s,0,sizeof(s)) ;        memset(p,0,sizeof(p)) ;        memset(next , 0, sizeof(next));        scanf("%s%s",p,s);        lenp = strlen(p);//*匹配串        lens = strlen(s);//*主串        kkk();        int ans = KMP();        cout<<ans<<endl;    }    return 0 ;}