题解报告:hdu 1686 Oulipo(裸KMP)
Problem Description
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
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
Sample Input
Sample Output
#include<cstdio>
#include<string.h>
const int maxn=1e4+;
const int maxm=1e6+;
char text[maxm],pattern[maxn];
int prefix[maxn],lena,lenb,num,t;
void get_prefix_table(){//处理模式串前缀表
int j=,pos=-;
prefix[]=-;
while(j<lenb){
if(pos==-||pattern[pos]==pattern[j])prefix[++j]=++pos;
else pos=prefix[pos];
}
}
void kmp_search(){
int i=,j=;
while(i<lena){
if(j==-||text[i]==pattern[j])i++,j++;
else j=prefix[j];
if(j==lenb)num++,j=prefix[j];//此时的j应退回到j前面子串的最长公共前后缀长度的位置进行下一次匹配,可以有相交的模式串
}
}
int main(){
while(~scanf("%d",&t)){
while(t--){
scanf("%s%s",pattern,text);
memset(prefix,,sizeof(prefix));
num=,lena=strlen(text),lenb=strlen(pattern);
get_prefix_table();
kmp_search();
printf("%d\n",num);
}
}
return ;
}
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