http://acm.hdu.edu.cn/showproblem.php?pid=1686

Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1677 Accepted Submission(s): 646


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

题意就是找模式串在母串中出现的次数，，，kmp算法。。。这里是从1开始，比较方便。。。

代码：

#include<iostream>
#include<cstdio>
#include<string.h>
#define N 10005
#define M 1000005
using namespace std;
 char a[N],s[M];
 int next[N];
 int n,m;
 void KMP_next()
 {     a[0]='#';//这里是从a+1开始输入，一定要注意，给a[0]赋值
       n=strlen(a);
       m=strlen(s);
       next[1]=0;
       int j=0;
       for(int i=2;i<n;i++)
       {
        while(j>0&&a[j+1]!=a[i])
            j=next[j];
            if(a[j+1]==a[i])
               j++;
               next[i]=j;
        }
 }
 int KMP_match()
 {   int sum=0,i=0,j=0;
      for(int i=0;i<m;++i)
      { while(j>0&&a[j+1]!=s[i])
            j=next[j];
            if(a[j+1]==s[i])
              j++;
              if(j==n-1) 
              { sum++;
                 j=next[j];
              }
      }
      return sum;
     }
     int main()
     {   int Case;
        scanf("%d",&Case);
         while(Case--)
         {    //getchar();
          scanf("%s%s",a+1,s);
             KMP_next();
              printf("%d\n",KMP_match());
             
                      } return 0;
         } 

这个超时的不知为什么，大牛看到时指点一下不胜感激，，，

代码：

#include<iostream>
#include<string>
#define  N 10005
using namespace std;
int next[N];
int  KMP_next(string &a,string &b)
{  int n=a.size();
   int m=b.size();
   int j=0,i=1,sum=0;
    next[0]=0;
    while(i<n)
    { if(a[j]==a[i])
        {  next[i]=j+1;
            j++;i++;
        }
      else{  if(j>0) j=next[j-1];
                else  next[i++]=0;
             }
    }
     i=0,j=0;
     while(i<m)
     {  if(a[j]==b[i])
       {  if(j==n-1) {sum++;j=next[j-1]; }
           else {j++;i++;}
       }
       else   { if(j>0) j=next[j-1];
                 else i++;
               }
       }
       return sum;
     }
     int main()
     {  int Case;
         cin>>Case;
         while(Case--)
         {  string a,b;
            cin>>a>>b;
            cout<<KMP_next(a,b)<<endl;
            }  return 0;
         }