poj 3461 Oulipo（KMP模板题）

Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 36903		Accepted: 14898

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

Source

/*
题意：给你两个字符串P,T让你判断P在T中出线了多少次
*/

/*
next数组就是表示以第i位为结尾的字符串，前缀后缀中最长相等字符串的长度
*/
#include<stdio.h>
#include<iostream>
#include<string.h>
#define M 1000010
using namespace std;
int next[M],sum,cur=0;
void makeNext(const char P[],int next[])
{
    int q,k;//q:模版字符串下标；k:最大前后缀长度
    int m = strlen(P);//模版字符串长度
    next[0]=0;//模版字符串的第一个字符的最大前后缀长度为0
    for (q=1,k=0;q<m;++q)//for循环，从第二个字符开始，依次计算每一个字符对应的next值
    {
        /*
        这里的while的作用：如果p[q]和p[k]相匹配了，那就说明前q字母组成的字符串最长的长度就是k了，如果没有匹配上的话，就看看前一个
        状态最长前后缀长度next[k-1]能不能和p[q]匹配一直这么找下去。
        */
        while(k>0&&P[q]!=P[k])//递归的求出P[0]···P[q]的最大的相同的前后缀长度k
        {
            k=next[k-1];          //不理解没关系看下面的分析，这个while循环是整段代码的精髓所在，确实不好理解
        }    
        if (P[q]==P[k])//如果相等，那么最大相同前后缀长度加1
        /*仔细想想如果这最后一个字母和遍历到的这个字母都一样了，那么最长的匹配块肯定就会多一个*/
        {
            k++;
        }
        next[q]=k;
    }
}
int kmp(const char T[],const char P[],int next[])
{
    int n,m;
    int i,q;
    n=strlen(T);
    m=strlen(P);
    makeNext(P,next);
    for(i=0,q=0;i<n;++i)
    {
        while(q>0&&P[q]!=T[i])
            q=next[q-1];
        /*这里的while就是参照next数组中的q位置的值，看看需要向前移动多少个单位*/
        if (P[q]==T[i])
        {
            q++;
        }
        if(q==m)//找到一个完整的字符串
        {
            cur++;
        }
    }    
}

int main()
{
    freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    char p[10005],t[1000005];
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s\n%s\n",&p,&t);
        cur=0;
        memset(next,0,sizeof next);
        makeNext(p,next);
        kmp(t,p,next);
        printf("%d\n",cur);
    }
    return 0;
}

来源：https://www.cnblogs.com/wuwangchuxin0924/p/5974511.html

标签

kmp