九度oj 1094

题目描述：

    Finding all occurrences of a pattern in a text is a problem that arises frequently in text-editing programs. 
    Typically,the text is a document being edited,and the pattern searched for is a particular word supplied by the user.  
    We assume that the text is an array T[1..n] of length n and that the pattern is an array P[1..m] of length m<=n.We further assume that the elements of P and  T are all alphabets(∑={a,b...,z}).The character arrays P and T are often called strings of characters.  
    We say that pattern P occurs with shift s in the text T if 0<=s<=n and T[s+1..s+m] = P[1..m](that is if T[s+j]=P[j],for 1<=j<=m).  
    If P occurs with shift s in T,then we call s a valid shift;otherwise,we calls a invalid shift. 
    Your task is to calculate the number of vald shifts for the given text T and p attern P.

输入：

   For each case, there are two strings T and P on a line,separated by a single space.You may assume both the length of T and P will not exceed 10^6. 

输出：

    You should output a number on a separate line,which indicates the number of valid shifts for the given text T and pattern P.

样例输入：

abababab abab

样例输出：

3

来源：

2006年上海交通大学计算机研究生机试真题

#include<iostream>#include<string.h>const int max=1000000+5;char p[max];char t[max];int next[max];using namespace std;void getnext(int size){     int i=0;     int j=-1;     next[0]=-1;     while(i<size)     {                  if(j==-1||t[i]==t[j])                  {                                       i++;                                       j++;                                       (t[i]==t[j])?next[i]=next[j]:next[i]=j;                                       }                                       else                                       j=next[j];                                       }                                       }                                       int kmp(int size1,int size2)                                       {                                           int i=0;                                           int j=0;                                           int ans=0;                                           while(i<size1)                                           {                                                         if(j==-1||p[i]==t[j])                                                         {                                                                              i++;                                                                              j++;                                                                              }                                                                              else                                                                              {                                                                                  j=next[j];                                                                                  }                                                                                  if(j==size2)                                                                                  {                                                                                              ans++;                                                                                              j=next[j];                                                                                              }                                                                                              }                                                                                                                                                                                            return ans;                                                                                              }                                                                                              int main()                                                                                              {                                                                                                  while(cin>>p>>t)                                                                                                  {                                                                                                                                                                                                                                    int l1=strlen(p);                                                                                                                  int l2=strlen(t);                                                                                                                  getnext(l2);                                                                                                                  cout<<kmp(l1,l2)<<endl;                                                                                                                  }                                                                                                                  }