POJ 3461 Oulipo （kmp模板题）

题目链接：POJ 3461

Describe:

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

题目大意：

若干样例，每个样例第一行给出一个单词，第二行给出一个字符串，要求输出，字符串中有多少个给定的单词。

解题思路：

比较裸的KMP，不过有一点需要根据kmp得到的next数组进行优化，在我们匹配完一次单词以后，如果所给单词从头匹配就会TLE，所以此时让下标j=next[j-1]，道理很简单，同样是根据前缀后缀，公共的部分已经匹配过，不需要重复匹配。

AC代码：

 

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
char str[1000010];   // 输入的字符串
char pattern[10010]; // 输入的单词，相当于模式串
int next[10010];
int n1,n2; // 两个串的长度
// 常规next数组
void getnext()
{
    memset(next,0,sizeof(next));
    int i,j;
    for(i=1,j=0; i < n2; )
    {
        if(pattern[i] == pattern[j]) {
            next[i] = next[i-1]+1;
            i++;j++;
        } else {
            if(j != 0)
            {
                j = next[j-1];
            } else next[i++] = 0;
        }
    }
}
// 常规kmp
int kmp()
{
    int ans = 0; // 记录出现次数
    int i,j;
    for(i=0,j=0; i <= n1; ) // 循环条件注意，我这种写法要有等号
    {
        if(j == n2)
        {
            j = next[j-1]; // 重点，根据next数组优化
            ans++;
            if(i == n1) break;
        }
        if(str[i] == pattern[j])
        {
            i++;
            j++;
        } else {
            if(j != 0) j = next[j-1];
            else i++;
        }
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",pattern,str);
        n1 = strlen(str);
        n2 = strlen(pattern);
        getnext();
        printf("%d\n",kmp());
    }
    return 0;
}