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.
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.
OutputFor 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 AVERDXIVYERDIANSample Output
1 3
0
題目意思:給你一個子串,求該子串在主串中出現的次數。
解題思路:利用kmp,不斷匹配,匹配到了計數器就加1,但是這個題我第一次寫超時了,因爲一次匹配成
功後,我就將n置爲0,讓主串倒退到上一個出現子串開頭的地方。
後來改進了一下,主串其實不用倒退,只需將子串倒退到next[n]的位置,即可!
#include<iostream>
#include<cstring>
using namespace std;
int const maxn=1000005;
char str[maxn],str1[maxn];
int next1[maxn];
int len,len1;
void sign()
{
int i,j;
next1[0]=-1;
i=-1,j=0;
int len1=strlen(str1);
while(j<len1)
{
if(i==-1||str1[i]==str1[j])
{
i++;
j++;
next1[j]=i;
}
else
{
i=next1[i];
}
}
}
int kmp()
{
int m=0,n=0,num=0;
int len=strlen(str);
int len1=strlen(str1);
sign();
while(m<len)
{
if(n==-1||str[m]==str1[n])
{
m++;
n++;
}
else
{
n=next1[n];
}
if(n==len1)
{
n=next1[n];
num++;
}
}
return num;
}
int main()
{
int T;
cin>>T;
while(T--)
{
scanf("%s",str1); //子串
scanf("%s",str); //主串
int num1=kmp();
cout<<num1<<endl;
}
return 0;
}