Oulipo
Time Limit: 1000MS Memory Limit: 65536K
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
题目大意:输入T组数据。每组数据两个字符串,你将第一个字符串在第二个字符串中出现的次数输出。如AZA在AZAZA中有2次,在AZAZAZA中有三次。
分析:典型的KMP问题。单模式匹配。
不懂KMP算法的请参考:
http://www.ruanyifeng.com/blog/2013/05/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm.html
#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int maxn = 1e4+5;
int next[maxn];
char str1[maxn];
char str2[1000005];
/*/////简单但是会超时
void MakeNext( char *p ){
int i,k;//Q记录字符串下标 k是最大前后相同缀长度
int m = strlen(p);//字符串长度
next[0] = -1;//第一个字符串的前后缀长度为0
for( i=1 , k=0 ; i<m ; i++ ){
while( k>0 && p[i] != p[k] ){
//求出p[0]-->p[i]的最大前后缀长度
k = next[k-1];
}
if( p[i] == p[k] ){
k++;
}
next[i] = k;
}
}
*/
void MakeNext( char *str ){
int n = strlen(str);
int i = 0, j = -1;
next[0] = -1;
while( i < n ){
if( j == -1 || str[i] == str[j] )
next[++i] = ++j;
else
j = next[j];
}
}
int kmp( char *p,char *str ){
int n = strlen(p);
int m = strlen(str);
int i = 0;
int j = 0;
int cnt = 0;
while( i<m && j<n ){
if( str[i] == p[j] || j == -1 ){
i++;
j++;
}
else{
j = next[j];
}
if( j == n ){
cnt++;
j = next[j];
}
}
return cnt;
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%s",str1);
scanf("%s",str2);
MakeNext( str1 );
int ans = kmp( str1,str2 );
printf("%d\n",ans);
}
return 0;
}