kmp(字符串匹配)

    kmp 算法，可以计算模式串是否在主串中出现，以及出现的位置。

    kmp_nextt  模式串的自我匹配， 

     j            1 2 3 4 5 6 7 8

     模式     a b a a b c a c

     nextt[j]  0 1 1 2 2 3 1 2

     get_nextval() 是 kmp_nextt 的优化

     具体例子：

     j               1 2 3 4 5

    模式         a a a a b

    nextt[j]      0 1 2 3 4

    nextval[j]  0 0 0 0 4

代码：

#include <stdio.h>#include <string.h>const int N = 10100;char a[N], b[N]; // 主串  a, 模式串 b。 int nextt[N]; //int nextval[N]; void kmp_nextt(int blen) // 模式串 b 的自我匹配 {int i = 1, j = 0;nextt[1] = 0;while(i < blen){if(j == 0 || b[i] == b[j]) {++i, ++j;nextt[i] = j;}else    j = nextt[j];} }void get_nextval(int blen) // 模式串自我匹配的优化 {int i = 1, j = 0;nextval[1] = 0;while(i < blen){if(j == 0 || b[i] == b[j]){++i, ++j;if(b[i] != b[j])nextval[i] = j;else    nextval[i] = nextval[j];}else{j = nextval[j];}} }int kmp(int alen, int blen, int pos)//求模式串 b , 在主串 a 第 pos 个字符之后的位置 {int  i = pos, j = 1;while(i <= alen && j <= blen){     if(j == 0 || a[i] == b[j])     {          ++i, ++j;  } else     j = nextt[j]; } if(j > blen) return i - blen; // 匹配成功返回一个位置 else return false;  } int main(){     scanf("%s %s", a+1, b+1);     // 主串 模式串 ， 下表都是从 1 开始       int alen = strlen(a+1); // 主串长度      int blen = strlen(b+1); // 模式串长度 kmp_nextt (blen); printf("%d\n", kmp (alen, blen, 1)); return 0; }

下面有几道题验证下代码：

验证下 kmp_nextt:

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 52257 Accepted: 21771

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcdaaaaababab.

Sample Output

143

题意：

就是问最多有几个循环节。最少一个，就是自身。

思路：

求nextt[b+1] 回到那里，blen - (nextt[blen+1]-1) 就是循环节的最短长度。

代码：

#include <stdio.h>#include <string.h>const int N = 1001000; char b[N];int nextval[N]; int nextt[N];void kmp_nextt(int blen) //模式串自我匹配 {int i = 1, j = 0;nextt[1] = 0;while(i <= blen){if(j == 0 || b[i] == b[j]) {++i, ++j;nextt[i] = j;}else    j = nextt[j];} }int main(){while(scanf("%s",b+1), b[1]!='.') // 下标从 1 开始 { int blen = strlen(b+1);  kmp_nextt(blen); // for(int i=1; i<=blen+1; i++)//   printf("%d ",nextt[i]); int ans = 1;   if(blen%(blen-(nextt[blen+1]-1)) == 0)    ans = blen / (blen - (nextt[blen+1]-1));     printf("%d\n",ans);}return 0;} 

验证下 kmp。

                                                          Oulipo

                                                               Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                      Total Submission(s): 16251    Accepted Submission(s): 6479

Problem 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

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

 

Sample Output

130

题意：上面哪个是模式串，下面的是主串，问模式串在主串中出现几次。

代码：

#include <stdio.h> #include <string.h>const int N = 1000100;char a[N], b[N]; // 主串  a, 模式串 b。 int nextt[N]; //void kmp_nextt(int blen){int i = 1, j = 0;nextt[1] = 0;while(i < blen){if(j == 0 || b[i] == b[j]) {++i, ++j;nextt[i] = j;}else    j = nextt[j];} }int ans = 0;int kmp(int alen, int blen, int pos)//求模式串 b , 在主串 a 第 pos 个字符之后的位置 {int  i = pos, j = 1;while(i <= alen && j <= blen){     if(j == 0 || a[i] == b[j])     {          ++i, ++j;         if(j > blen || j == blen && a[i] == b[j])  //  每当成功匹配 ，结果 +1          {          ans ++;          j = nextt[j]; }  } else     j = nextt[j]; } if(j > blen) return i - blen;else return false;  } int main(){ int n; scanf("%d ",&n); while(n--) { scanf("%s %s", b+1, a+1);        int alen = strlen(a+1); // 主串长度         int blen = strlen(b+1); // 模式串长度        ans = 0;    kmp_nextt (blen);    kmp(alen, blen, 1);    printf("%d\n",ans); } return 0;}

get_nextval 找到相应题会补上。