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POJ3461 Oulipo 【KMP】
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 22295 | Accepted: 8905 |
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
奇了怪了，最近做题coding时觉得很顺利，但是debug的时候各种问题，今天第31行那里很久才发现漏洞 >_<|||
题意：给定一个模式串和主串，求主串中有多少个模式串（包含重叠）。
题解：这题是KMP的入门题了，只需要从左到右扫描一遍即可得出答案，next数组的意义为：若模式串第j个字符与主串第i个字符不匹配那么模式串指针j必须要回溯到一个位置重新与主串中的第i个字符开始匹配，next就是用来记录这个回溯的位置的，
#include <stdio.h>
#define maxn 10002
#define maxN 1000002
char str1[maxN], str2[maxn];
int next[maxn];
void getNext()
{
int i = 0, j = -1;
next[0] = -1;
while(str2[i]){
if(j == -1 || str2[i] == str2[j]){
++i; ++j;
if(str2[i] == str2[j]) next[i] = next[j];
else next[i] = j;
}else j = next[j];
}
}
int KMP()
{
int ans = 0;
getNext();
int i = 0, j = 0;
while(str1[i]){
if(j == -1 || str1[i] == str2[j]){
++i; ++j;
}else j = next[j];
if(j != -1 && !str2[j]){ //Attention!!!don't forget "j != -1"!
++ans; j = next[j];
}
}
return ans;
}
int main()
{
//freopen("stdin.txt", "r", stdin);
int t;
scanf("%d", &t);
while(t--){
scanf("%s%s", str2, str1);
printf("%d\n", KMP());
}
return 0;
}