KMP演算法的next、next value陣列程式碼實現及POJ3461

kewlgrl發表於2015-08-12
KMP演算法陣列


昨天中午弄懂了陣列的手工計算方法之後，根據書上例題解出了一道KMP演算法的匹配題。

我用了next 和nextval兩種解決方法，其實就是陣列實現的程式碼片不同。

w表示給定的模式字串

next陣列程式碼實現如下：

int next[maxw],j=0,i;
next[0]=-1;
next[1]=0;
for(i=2; i<=strlen(w); ++i)
{
    while(j>=0&&w[j]!=w[i-1])
        j=next[j];
    next[i]=++j;
}


nextval陣列程式碼實現如下：

int nextval[maxw],i=0,j=-1;
nextval[0]=-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];  
}


下面貼一道POJ上的例題：

Oulipo
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 29204
Accepted: 11704

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

這道題就是求匹配過程中子串在主串中出現了多少次。

next版：

#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;
}

nextval版：

#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;
}


比較記憶體佔用和執行時長，發現就這道題而言，nextval的用時要少但是記憶體佔用較多。

總之，nextval是對next優化改進後的方法，效率會提高。

我對KMP演算法的初步學習大概就是這麼多認識，當然是木有BF演算法那麼好理解，但是KMP又快又好用阿~~

希望我能就這麼堅持下去吧，即使腦子木有人家那麼靈活但是如果多花花時間能弄出來我也是挺開心噠~~

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