POJ-3461 Oulipo-匹配的字元有幾個（KMP演算法）

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 wordW and a text T, count the number of occurrences of W inT. All the consecutive characters of W must exactly match consecutive characters ofT. Occurrences may overlap.

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 stringW).
• One line with the text T, a string over {'A', 'B','C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.

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 wordW in the text T.

/*    
*Copyright (c)2015,煙臺大學計算機與控制工程學院    
*All rights reserved.      
*作    者：單昕昕    
*完成日期：2015年8月16日    
*版 本 號：v1.0        
*/
//nextval陣列
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxw=10000+10;
const int maxt=1000000+10;

int match(char w[],char s[],int next [])
{
    int cnt=0,p=0,cur=0,slen,wlen;
    slen=strlen(s);
    wlen=strlen(w);
    while(cur<slen)
    {
        if(s[cur]==w[p])
        {
            ++cur;
            ++p;
        }
        else if(p>=0)
        {
            p=next[p];
        }
        else
        {
            ++cur;
            p=0;
        }
        if(p==wlen)
        {
            ++cnt;
            p=next[p];
        }
    }
    return cnt;
}

int main()
{
    int loop;
    scanf("%d",&loop);
    while(loop--)
    {
        char w[maxw],t[maxt];
        scanf("%s%s",w,t);
        int next[maxw],p=0,cur;
        next[0]=-1;
        next[1]=0;
        for(cur=2; cur<=strlen(w); ++cur)
        {
            while(p>=0&&w[p]!=w[cur-1])
                p=next[p];
            next[cur]=++p;
        }
        printf("%d\n",match(w,t,next));
    }
    return 0;
}

/*    
*Copyright (c)2015,煙臺大學計算機與控制工程學院    
*All rights reserved.      
*作    者：單昕昕    
*完成日期：2015年8月16日    
*版 本 號：v1.0        
*/
//nextval陣列
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxw=10000+10;
const int maxt=1000000+10;

int match(char w[],char s[],int next [])
{
    int cnt=0,p=0,i=0,slen,wlen;
    slen=strlen(s);
    wlen=strlen(w);
    while(i<slen)
    {
        if(s[i]==w[p])
        {
            ++i;
            ++p;
        }
        else if(p>=0)
        {
            p=next[p];
        }
        else
        {
            ++i;
            p=0;
        }
        if(p==wlen)
        {
            ++cnt;
            p=next[p];
        }
    }
    return cnt;
}

int main()
{
    int loop;
    scanf("%d",&loop);
    while(loop--)
    {
        char w[maxw],t[maxt];
        scanf("%s%s",w,t);
        int nextval[maxw],i,j;
        i=0;
        nextval[0]=-1;
        j=-1;
        while(i<strlen(w))
        {
            if(j==-1||w[i]==w[j])
            {
                ++i;
                ++j;
                if (w[i]!=w[j])
                    nextval[i]=j;
                else
                    nextval[i]=nextval[j];
            }
            else
                j=nextval[j];
        }
        printf("%d\n",match(w,t,nextval));
    }
    return 0;
}