hdu1686字符串kmp

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.

InputThe 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. 
OutputFor 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：还是找最大匹配，但是这次可以重复
题解：刚开始用的上一题的第一个方法，直接暴力遍历字符串结果tle，发现可以利用字符串kmp本身的特点处理

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
 
using namespace std;
 
const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f3f;
 
int Next[N],slen,plen,res;
string str,ptr;
 
void getnext()
{
    int k=-;
    Next[]=-;
    for(int i=;i<=slen-;i++)
    {
        while(k>-&&str[k+]!=str[i])k=Next[k];
        if(str[k+]==str[i])k++;
        Next[i]=k;
    }
}
void kmp()
{
    int k=-;
    for(int i=;i<plen;i++)
    {
        while(k>-&&str[k+]!=ptr[i])k=Next[k];
        if(str[k+]==ptr[i])k++;
        if(k==slen-)
        {
            res++;
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie();
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    int t;
    cin>>t;
    while(t--){
        cin>>str>>ptr;
        slen=str.size();
        plen=ptr.size();
        getnext();
        res=;
        kmp();
        cout<<res<<endl;
    }
    return ;
}