HDU1686(KMP多次匹配)

Oulipo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12783    Accepted Submission(s): 5025


Problem 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:求字符串在第二个字符串匹配次数
解题思路:KMP加上在判断上进行一个变化。主要在于KMP()中while的一个变化
#include  
#include
#include
#include  
#include  
#include  
#include 
#include
#include 
#include  
#include 
#include
using namespace std;
string str1, str2;
long long int nxt[1000008], n, m, T;
char b[1000008], a[1000008];
void get_next()
{
	long long int i, j;
	i = 1;
	j = 0;
	nxt[1] = 0;
	while (i<=n)
	{

		if (a[i] == a[j]||j==0)
		{
			i++; j++;
			nxt[i] = j;
		}
		else
		{
			j = nxt[j];
		}
	}
}
long long int kmp()
{
	long long int i, j, sum;
	sum = 0;
	i = 0;
	j = 0;
	while (j<=m)
	{
		if (a[i] == b[j]||i==0)
		{
			i++;
			j++;
		}
		else
		{
			if (i>n)
			{
				sum++;
				i = nxt[i];
			}
			else
			{
				i = nxt[i];
			}
		}
	}
	if (i > n)
	{
		sum++;
	}
	return sum;
}
int main()
{
	string str1, str2;
	long long int i;
	cin >> T;
	while (T--)
	{
		memset(a, 0, sizeof(a));
		memset(b, 0, sizeof(b));
		cin>> str1;
		n = str1.length();
		for (i = 0; i < n; i++)
		{
			a[i + 1] = str1[i];
		}
		cin >> str2;
		m = str2.length();
		for (i = 0; i < m; i++)
		{
			b[i + 1] = str2[i];
		}
		if (m < n)
		{
			cout << 0 << endl;
			continue;
		}
		memset(nxt, 0, sizeof(nxt));
		get_next();
		cout << kmp() << endl;
	}
}


 

你可能感兴趣的:(ACM,模板题,ACM算法,日常总结,KMP)