HDU 1686 Oulipo 求大串中最多可匹配多少个小串(kmp)

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): 6098    Accepted Submission(s): 2448

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

 

 

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <cstring>
using namespace std;
int n,m,nxt[10005],kk,t;
char b[10005],a[1000005];
///此题在基础的kmp上加了多次匹配。
///就意味着我们在匹配完一次字串后，要跳到最适合的位置,继续查找
///继续利用kmp的思想。某些位置已经匹配过，就不要匹配了。
///       xxxxxxxabbaab*xxxxxx
///              abbaaba
//我们跳跃之后的位置 abbaaba  而跳跃的位置与next数组有关
//kmp中主串的位置都没有被调动，只是next数组的下标被调动（自己写的代码乱动了，我又卖萌了。。。）
void buildnxt()
{
   int j,k;
   m=strlen(b);
    nxt[0]=-1;
    j=0;k=-1;
    while(j<m)
    {
        if((k==-1)||b[j]==b[k])
        {
            j++;
            k++;
            nxt[j]=k;
        }
        else k=nxt[k];
    }
}
int kmp()
{
    int k=0,l=0,cou=0;
    n=strlen(a);
    /*int ans=m,kk=nxt[m];///ans在字串中下标，和起点距离ans+1
    while(1)
    {
        if(kk!=0&&kk!=-1) {ans=kk;kk=nxt[ans];}
        else break;
    }///要找最小的跳跃点,所以从next尾端返回去找到首个非负值。
额 这个想法是没错，但是时间上还是不够优化。
对与最小跳跃点的话也就是中间跳跃点，最小跳跃点对应的字符串匹配失败。没必要要该点匹配。
于是我们最省事的做法还是直接往前跳一步，有可能匹配成功。
*/
    while(k<n)
    {
        if((l==-1)||a[k]==b[l])
        {
            k++;
            l++;
        }
        else l=nxt[l];
        if(l==m)
        {
            cou++;
            /*if(kk==0) continue;///如果是尾端next数组是0的话，主串中匹配的子串中没有重复。
            ///也就是说在匹配的主串中，没有可以跳跃的点。
            if(k==n-1) break;///如果k已经是主串末尾了，就不能还有继续可以匹配的字串了。
            k=k-l+ans;///k-l(起点)+ans
            l=0;*/
l=nxt[l];//next跳到次大子串点重新匹配，跳过已经匹配好的部分
        }
    }
    return cou;
}
int main()
{
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        gets(b);
        gets(a);
        memset(nxt,0,sizeof(nxt));
        buildnxt();
        printf("%d\n",kmp());
    }
    return 0;
}