Oulipo （poj3461

Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29759		Accepted: 11986

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

Source

BAPC 2006 Qualification

题意：两个字符串，在第二个字符串中可以找到多少个含有第一个字符串的区间……

kmp能干嘛，next是干嘛的。

按照正常思维一个个查的话，效率太低，于是伟大的前辈们就创造出了针对如何快速查找字符串的高大上的kmp。

 

理解：    因为每次要查找的时候都会对字符串从头开始查找，但是如果我们已经知道在查找字符串当前位置的时候，有部分前缀和后缀是相等的，但是当前位置又不相等，

那下次查找的时候就不用从头开始找啦，因为前缀和后缀相等，所以前缀和后缀相等的长度就不用查询了，就从已经查找到的位置减去前缀和后缀相等的长度开始查询就行了。

那么问题又转化成了怎么求字符串当前位置前缀和后缀相等的长度。用数组next存储，next[j] = k,意味着：p[0, k-1] = p[j-k, j-1].

 

 

http://www.cnblogs.com/dolphin0520/

因此KMP算法的关键在于求算next[]数组的值，即求算模式串每个位置处的最长后缀与前缀相同的长度， 而求算next[]数组的值有两种思路，第一种思路是用递推的思想去求算，还有一种就是直接去求解。

1.按照递推的思想：

根据定义next[0]=-1，假设next[j]=k, 即P[0...k-1]==P[j-k,j-1]

1)若P[j]==P[k]，则有P[0..k]==P[j-k,j]，很显然，next[j+1]=next[j]+1=k+1;

2)若P[j]!=P[k]，则可以把其看做模式匹配的问题，即匹配失败的时候，k值如何移动，显然k=next[k]。

因此可以这样去实现：

 void getNext(char *p,int *next)
 {
     int j,k;
     next[]=-;
     j=;
     k=-;
     while(j<strlen(p)-)
     {
         if(k==-||p[j]==p[k])    //匹配的情况下,p[j]==p[k]
         {
             j++;
             k++;
             next[j]=k;
         }
         else                   //p[j]!=p[k]
             k=next[k];
     }
 }

问题： next应该到数组长度……

优化一点：

 void getNext(char *p,int *next)
 {
     int j,k;
     int lastIndex=strlen(p)-;
     next[]=-;
     j=;
     k=-;
     while(j < lastIndex)
     {
         if(k==-||p[j]==p[k])    //匹配的情况下,p[j]==p[k]
         {
             ++j;
             ++k;

             //若p[j]==p[k]，则需要修正
             if(p[j]==p[k])
                 next[j]=next[k];
             else
                 next[j]=k;
         }
         else                   //p[j]!=p[k]
             k=next[k];
     }
 }

2.直接求解方法

 void getNext(char *p,int *next)
 {
     int i,j,temp;
     for(i=;i<strlen(p);i++)
     {
         if(i==)
         {
             next[i]=-;     //next[0]=-1
         }
         else if(i==)
         {
             next[i]=;      //next[1]=0
         }
         else
         {
             temp=i-;
             for(j=temp;j>;j--)
             {
                 if(equals(p,i,j))
                 {
                     next[i]=j;   //找到最大的k值
                     break;
                 }
             }
             if(j==)
                 next[i]=;
         }
     }
 }

 bool equals(char *p,int i,int j)     //判断p[0...j-1]与p[i-j...i-1]是否相等
 {
     int k=;
     int s=i-j;
     for(;k<=j-&&s<=i-;k++,s++)
     {
         if(p[k]!=p[s])
             return false;
     }
     return true;
 }

本题代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>

 using namespace std;

 char w[], t[];
 int next[];

 void getNext()
 {
     int j, k;
     int i = strlen(w);
     j = ;
     k = -;
     next[] = -;

     while(j < i)
     {
         if(k == - || w[j] == w[k])
         {
             j++;
             k++;
             next[j] = k;
         }
         else
             k = next[k];
     }
 }

 int main()
 {
     int c;
     scanf("%d", &c);
     while(c--)
     {
         memset(next, , sizeof(next));
         scanf("%s", w);
         scanf("%s", t);
         getNext();
         int i, j, l1, l2;
         l1 = strlen(w);
         l2 = strlen(t);
         i = j = ;
         int sum = ;
         while(i < l1 && j < l2)
         {
             if(w[i] == t[j] || i == -)
             {
                 i++;
                 j++;
             }
             else
                 i = next[i];            

             if(i == l1)
             {
                 sum++;
                 i = next[i];
             }
         }
         printf("%d\n", sum);
     }
     return ;
 }