POJ3461_Oulipo_KMP_求重复子串的个数_可重叠

题意：

给母串str，和子串w，求在str中最多有几个w，w可以相互重叠

比如

str：ABABABA

w：ABA

ans=3

题解：

裸的KMP算法，只是这时候不是返回子串的位置，而是重复KMP遍历完整个串求个数

原题：

Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16626		Accepted: 6656

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 wordW and a text T, count the number of occurrences of W inT. All the consecutive characters of W must exactly match consecutive characters ofT. 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 stringW).
• 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 wordW in the text T.

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Run ID	User	Problem	Result	Memory	Time	Language	Code Length	Submit Time
11776963	chengtbf	3461	Accepted	1192K	141MS	C++	1043B	2013-07-13 17:08:24

代码：

#include
#include
#define N 1000005

char str[N],w[10005];
int next[10005];
int n;//n为母串长度
int m;//m为子串长度
int count;

void get_next()
{
	int pos=2;int cnd=0;
	next[0]=-1;next[1]=0;
	while (pos<=m)
	{
		if (w[pos-1]==w[cnd])
		{
			cnd++;
			next[pos]=cnd;
			pos++;
		}
		else if(cnd>0)
		{
			cnd=next[cnd];
		}
		else
		{
			next[pos]=0;pos++;
		}
	}
}

void search()
{
	int pos,i;
	pos=0;
	i=0;
	while (pos+i<=n-1)
	{
		if (w[i]==str[pos+i])
		{
			if (i==m-1)
			{
				count++;
				pos=pos+i-next[i];//注意这里，pos++会超时，要优化成这个，表示pos到pos+i-next[i]之间的匹配一定不成功，由next[]数组含义可得
					if(next[i]>-1)i=next[i];//这里，也要注意，我第一次就是这里TLE的，没控制好死循环了，
											//这里也跟下面i的变化一模一样，直接i=0也会超时，优化一下，因为前next[i]个字符一定已经匹配成功了，
											//直接从next[i]开始继续向后匹配

					else i=0;//但是由于next[0]=-1，所以有可能因此死循环，所以要判断一下，此时取i=0；
			}
			else
			{
				i++;
			}
		}
		else
		{
			pos=pos+i-next[i];
			if (next[i]>-1)
			{
				i=next[i];
			}
			else
			{
				i=0;
			}
		}
	}
}

int main()
{
	int t;
	char temp;
	scanf("%d",&t);
	while (t--)
	{
		count=0;
		scanf("%c",&temp);
		scanf("%s",w);
		scanf("%s",str);
		n=strlen(str);
		m=strlen(w);
		get_next();
		search();
		printf("%d\n",count);

	}
	return 0;
}