題目
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.
代碼:
#include <iostream>
#include <string.h>
#include <stdio.h>
#define a 10000001
using namespace std;
int next[a],cnt;
char s[a],p[a];
void getnext(int len)
{
next[0]=-1;
int j=0;
int k=-1;
while(j<len-1)
{
if((k==-1)||(p[j]==p[k]))
{
j++;
k++;
if(p[j]!=p[k])
next[j]=k;
else
next[j]=next[k];
}
else
k=next[k];
}
}
void kmp(int slen,int plen)
{
int i=0,j=0;
getnext(plen);
while (i<slen)
{
if(j==-1||s[i]==p[j])
{
i++;
j++;
}
else
{
j=next[j];
}
if(j==plen)
{
cnt++;
j=next[j];
}
}
}
int main()
{
int n,slen,plen;
cin>>n;
getchar();
while(n--)
{
gets(p);
gets(s);
slen=strlen(s);
plen=strlen(p);
cnt=0;
kmp(slen,plen);
cout<<cnt<<endl;
}
return 0;
}
知識點:
KMP
http://blog.csdn.net/v_july_v/article/details/7041827