标签：kmp
原题http://poj.org/problem?id=3461
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 23987 | Accepted: 9613 |
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:
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
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
using namespace std;
#define MAXN1 10000 + 10
#define MAXN2 1000000 + 10
char a[MAXN1];
char b[MAXN2];
int next[MAXN1];
int len1;
int len2;
//int count;
void Getnext(){
int i = 0;
int j = -1;
next[0] = -1;
while(i < len1){
if(j==-1 || b[i]==b[j]){
i++;
j++;
next[i] = j;
}
else{
j = next[j];
}
}
}
/*int KMP(int s,int count){
if(len2-s < len1){
return count;
}
int i = s;
int j = 0;
while(i<len2 && j<len1){
if(j==-1 || b[i]==a[j]){
i++;
j++;
}
else{
j = next[j];
}
}
if(j == len1){
count++;
KMP(i-1,count);
}
else{
KMP(s+1,count);
}
}*/
int KMP(){
int count = 0;
int i = 0;
int j = 0;
for(i=0;i<len2;){
if(b[i]==a[j] || j==-1){
i++;
j++;
if(j == len1){
j = next[j];//慢慢体会这里。相信你可以的。
count++;
}
}
else{
j = next[j];
}
}
return count;
}
int main(){
int T,i;
while(~scanf("%d",&T)){
while(T--){
//count = 0;
scanf("%s%s",a,b);
len1 = strlen(a);
len2 = strlen(b);
Getnext();
//int mark = 0;
//while(len2-mark+1 >= len1){
// int mark = KMP(mark);
//}
if(len1 == 1){
int mark = 0;
for(i=0;i<len2;i++){
if(a[0] == b[i]){
mark++;
}
}
printf("%d\n",mark);
continue;
}
else{
int ans = KMP();
printf("%d\n",ans);
}
}
}
return 0;
}
标签：kmp
原文地址：http://blog.csdn.net/zcr_7/article/details/39076331
