学了这个算法，思路倒是不难懂，但是看了代码后却很绕，特别晕，一直有一个地方不太懂，现在才颇有感受。
next[0]为什么要赋值-1，以及这样的话char数组与next不就不对应了吗？
比如a[3]对应的next数组便是next[4]，但是跳跃的时候却是用next[3]
直到后来我准备自己实施代码时发现上面的做法太棒了
举个例子
数组1 ： a b c a b y x
对应的next数组：-1 0 0 0 1 2 0 0
数组2 ：abcabcabyx
如图，当对应到第5个元素时（对应数组的下标），数组1为y，数组2为c，不匹配，于是第一个数组的下标j = next[j]，即j = next[5] = 2，所以第一个数组下标就跳到第2个元素，也就是c了，c再与数组2的c匹配，完美！
如果不加上next[0] = -1会变成怎么样？那么每次匹配不成功，必须j = next[j-1]，而且无法判断是否已经回到了首元素。
附上刚刷的一道例题：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 46057 | Accepted: 18369 |
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：
找出第一个字符串在第二个字符串中出现次数。
附上AC代码：
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <stack>
#include <queue>
using namespace std;
#define ll long long
char a[11111];
int next[11111];
char b[1111111];
int lena;
int lenb;
int sum = 0;
void getnext()
{
int i,j;
next[0] = j = -1;
i = 0;
while(i < lena)
{
while(j != -1 && a[i] != a[j])
j = next[j];
next[++i] = ++j;
}
}
void kmp()
{
int i,j;
j = 0;
i = 0;
while(i < lenb)
{
while(j != -1 && a[j] != b[i])
j = next[j];
i++;j++;
if(j >= lena)
{
sum++;
j = next[j];
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
sum = 0;
scanf("%s%s",a,b);
lena = strlen(a);
lenb = strlen(b);
getnext();
kmp();
/*for(int i = 0;i <= len;i++)
cout << next[i] << ' ';
cout << endl;*/
printf("%d\n", sum);
}
return 0;
}