POJ3461 Oulipo

Oulipo

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 33738 Accepted: 13634

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

这是一道裸题，只是修改下，当完成一次匹配时，还进行跳转，这样就可以求出结果了

#include<iostream>#include<cstdio>#include<cstring>#define MAXN 100010using namespace std;void get_next(char* sub,int len_sub,int* next){    int i = 1;    int j = 0;    next[0] = -1;    while(i<len_sub)    {        if(j==-1||sub[i]==sub[j])        {           i++;           j++;           if(sub[i]!=sub[j]) next[i] = j;  //只有前后两个字符不相同时才会按前面的算法来计算next， 即next[i] = j           else next[i] = next[j];        //当前后两个字符相同时，进行向前找，能够减少比较的次数        }      else j = next[j];     }}inline int kmp(char *s,int len_s,char *sub,int len_sub,int* next){    int i=0,j=0,cot=0;    while(i<len_s && j<len_sub)    {        if(j==-1 || s[i]==sub[j])//第一个字符都不匹配或者字符相等的情况        {            i++;            j++;        }      else j=next[j];  //i指针不回溯,j指针      if(j==len_sub) {cot++;j=next[j];}    }    return cot;}int main(){    int n;    scanf("%d",&n);    while(n--)    {        int next[MAXN];        char s[MAXN*10],sub[MAXN];        memset(s,0,sizeof(s));        memset(sub,0,sizeof(sub));        memset(next,0,sizeof(next));        scanf("%s",sub);        scanf("%s",s);        int len_s=strlen(s),len_sub=strlen(sub);        get_next(sub,len_sub,next);        int ans=kmp(s,len_s,sub,len_sub,next);        printf("%d\n",ans);    }    return 0;}