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：用字符串1做模式串，字符串2做匹配串，输出匹配串的总数，允许重叠匹配
思路：kmp模板的套用，我自己思维没转过弯，所以wrong了一次，看了一下别人的题解才知道，允许重叠匹配的话，当匹配到模式串的末尾时，不能直接让j从0重新开始，而是利用next数组，让j回到最大匹配坐标。
#include<cstring>
#include<cstdio>
#define N 1100000
#define M 11000
char s2[M],s1[N];
int NEXT[M];
void getNext(char s2[],int next[])
{
int i,j,k;
k = -1;
j = 0;
next[0] = -1;//划重点！！！已经n多次忘记初始化了！！
while(s2[j]!='\0')
{
while(k!=-1&&s2[j]!=s2[k])
k = next[k];
k++;
j++;
if(s2[k]!=s2[j])
next[j] = k;
else
next[j] = next[k];
}
return;
}
void kmp(char s1[],char s2[],int next[])
{
int i,j,count;
i = j = count = 0;
while(s1[i]!='\0')
{
while(j!=-1&&s1[i] != s2[j])
j = next[j];
j++;
i++;
if(s2[j]=='\0')
{
j = next[j];
count ++;
}
}
printf("%d\n",count);
return;
}
int main()
{
int t;
scanf("%d",&t);
while(t --)
{
scanf("%s",s2);
scanf("%s",s1);
getNext(s2,NEXT);
kmp(s1,s2,NEXT);
}
return 0;
}