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: 65536K |
Total 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
abcd
aaaa
ababab
.Sample Output
1
4
3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
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
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
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;
}
