poj3461—Oulipo（查找子串在目标串中出现次数）

题目链接：传送门

Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44522		Accepted: 17847

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

直接上代码吧

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>

using namespace std;

const int N = 1005000;
const int M = 10100;

//poj3461
//计算子串在目标串中出现的次数

int nt
,pLen;
char s
,p[M];

void GetNext(char* p)
{
int pLen = strlen(p);
nt[0] = -1;
int k = -1,j = 0;
while (j < pLen - 1){
//p[k]表示前缀，p[j]表示后缀
if (k == -1 || p[j] == p[k]){
++k;
++j;
if (p[j] != p[k])
nt[j] = k; //之前只有这一行
else
//因为不能出现p[j] = p[ nt[j ]]
//所以当出现时需要继续递归，k = nt[k] = nt[nt[k]]
nt[j] = nt[k];
}else{
k = nt[k];
}
}
}

int KMP(char* s, char* p)
{
int i = 0,j = 0,ans = 0;
int sLen = strlen(s);
while (i < sLen){
//找到一处匹配
if(j == pLen){
ans++;
j = nt[pLen];
}
//如果j = -1,或
cd3c
者当前字符匹配成功(即S[i] == P[j]),都令i++，j++
if (j == -1 || s[i] == p[j]){
i++;
j++;
}else{
//如果j != -1,且当前字符匹配失败(即S[i] != P[j]),则令 i 不变，j = nt[j]
//nt[j]即为j所对应的nt值
j = nt[j];
}
}
if(j == pLen){
ans++;
}
return ans;
}

int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf(" %s",p);
scanf( "%s",s);
pLen = strlen(p);
p[pLen] = '#';
GetNext(p);

printf("%d\n",KMP(s,p));
}
return 0;
}