KMP算法详解
本篇文章相当于博主自己写的一个笔记
对于初学者来讲请仔细阅读如下网址
从头到尾彻底理解KMP（2014年8月22日版）
KMP算法的实现离不开一个重要的数组next[]数组
关于next[]数组的建立有两种方法一种next[0]=-1,另一种next[0]=0;
对于一些初学者来讲很容易分不清楚，下面先来介绍next[0]=-1的一种
主要是写一个代码模板，因为这性质是一篇笔记；
void GetNext(char *p,ll next[])
{
ll plen=strlen(p);//plen所求为p的长度，注意它比实际的位置小一
int i=0,j=-1;//这里i是一直向后移动,j与next[]有联系
next[0]=-1;//初始时j与next[0]的值相等
while(inext[0]=0的情况尽量还是用上面哪一种来写
void GetNext(char *p,ll next[])
{
ll plen=strlen(p);//plen所求为p的长度，注意它比实际的位置小一
int i=1,j=0;//这里i是一直向后移动,j与next[]有联系
next[1]=0;//初始时j与next[0]的值相等
while(iKMP算法应用于第一种
int KmpSearch(char* s, char* p)
{
int i = 0;
int j = 0;
int sLen = strlen(s);
int pLen = strlen(p);
while (i < sLen && j < pLen)
{
//①如果j = -1，或者当前字符匹配成功（即S[i] == P[j]），都令i++，j++
if (j == -1 || s[i] == p[j])
{
i++;
j++;
}
else
{
//②如果j != -1，且当前字符匹配失败（即S[i] != P[j]），则令 i 不变，j = next[j]
//next[j]即为j所对应的next值
j = next[j];
}
}
if (j == pLen)
return i - j;
else
return -1;
}
KMP算法应用于第二种
待定
KMP算法求一些问题，
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
#include
#include
#include
#include
using namespace std;
typedef long long ll;
char a[1000010],b[10010];
ll nextt[10010];
void get_next()
{
nextt[0]=-1;
int blen=strlen(b);
int i=0,j=-1;
while(i>T;
while(T--)
{
scanf("%s",b);
scanf("%s",a);
KMP();
}
return 0;
}
一块花布条，里面有些图案，另有一块直接可用的小饰条，里面也有一些图案。对于给定的花布条和小饰条，计算一下能从花布条中尽可能剪出几块小饰条来呢？
Input
输入中含有一些数据，分别是成对出现的花布条和小饰条，其布条都是用可见ASCII字符表示的，可见的ASCII字符有多少个，布条的花纹也有多少种花样。花纹条和小饰条不会超过1000个字符长。如果遇见#字符，则不再进行工作。
Output
输出能从花纹布中剪出的最多小饰条个数，如果一块都没有，那就老老实实输出0，每个结果之间应换行。
Sample Input
abcde a3
aaaaaa aa
Sample Output
0
3
#include
#include
#include
#include
using namespace std;
char s[1010],p[1010];
int nextt[1010];
void GetNext()
{
int i=0,j=-1;
nextt[0]=-1;
int plen=strlen(p);
while(i 关于next的优化算法
void GetNext(char *p,ll next[])
{
ll plen=strlen(p);//plen所求为p的长度，注意它比实际的位置小一
int i=0,j=-1;//这里i是一直向后移动,j与next[]有联系
next[0]=-1;//初始时j与next[0]的值相等
while(i