POJ3461：Oulipo（MP，KMP裸题）

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本文介绍了一种改进的字符串匹配算法——KMP算法，通过预处理模式串生成失配指针，避免了传统BF算法在遇到不匹配时重复比较的问题，提高了搜索效率。

Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 39112		Accepted: 15748

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

BAPC 2006 Qualification

题意：求一个串在另一个串中出现的次数。

思路：

①MP算法，区别于BF算法那样的暴力循环，当遇到不匹配的字符时，不需要回到模式串的开头重新匹配，只需从next处继续即可，如下图：遇到a和b不等，但黑色部分A和B是一样的话，直接从A的右边处继续匹配，next数组需要预处理。

# include <stdio.h>
# include <string.h>
char s[10001], t[1000001];
int next[10001];//失配指针
int slen, tlen;

void init()
{
    memset(next, 0, sizeof(next));
    next[0] = next[1] = 0;
    for(int i=1; i<slen; ++i)
    {
        int j = next[i];
        while(j && s[i] != s[j]) j = next[j];
        next[i+1] = s[i]==s[j]?j+1:0;
    }
}

int kmp()
{
    int ans = 0, j = 0;
    for(int i=0; i<tlen; ++i)
    {
        while(j && t[i] != s[j]) j = next[j];
        if(t[i] == s[j]) ++j;
        if(j == slen)
        {
            ++ans;//计数器
            j = next[j];
        }
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",s,t);
        slen = strlen(s);
        tlen = strlen(t);
        init();
        printf("%d\n",kmp());
    }
    return 0;
}

②KMP实现：区别于MP，多开一个数组用于优化，失配指针更精确。

# include <stdio.h>
# include <string.h>
char s[10001], t[1000001];
int next[10001];//优化后的失配指针
int next2[10001];//原来的失配指针
int slen, tlen;

void init()
{
    memset(next, 0, sizeof(next));
    memset(next2, 0, sizeof(next2));
    next[0] = next[1] = 0;
    next2[0] = next2[1] = 0;
    for(int i=1; i<slen; ++i)
    {
        int j = next2[i];
        while(j && s[i] != s[j]) j = next[j];
        next2[i+1] = next[i+1] = s[i]==s[j]?j+1:0;
        if(next[i+1]==j+1 && s[i+1]==s[j+1])//这种情况下，指针可以继续往前跳。
            next[i+1] = next[j+1];
    }
}

int kmp()
{
    int ans = 0, j = 0;
    for(int i=0; i<tlen; ++i)
    {
        while(j && t[i] != s[j]) j = next[j];
        if(t[i] == s[j]) ++j;
        if(j == slen)
        {
            ++ans;//计数器
            j = next[j];
        }
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",s,t);
        slen = strlen(s);
        tlen = strlen(t);
        init();
        printf("%d\n",kmp());
    }
    return 0;
}