Oulipo POJ - 3461 （KMP求s中有多少个p（两种题型））

                                              Oulipo POJ - 3461 

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：T组数据，下面分别为p，s，求s中有多少个p

思路：KMP

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<utility>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#define maxn1 1000005
#define maxn2 10005
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define E 1e-8
#define mod 1000000007
#define P pair<int,int>
using namespace std;

int n,m;
char s[maxn1];
char p[maxn2];
int nexts[maxn2];

void GetNexts()
{
    int j=0,k=-1;
    nexts[0] = -1;
    while(j<m){
        if(k==-1||p[j]==p[k]){
            j++;
            k++;
            nexts[j] = k;
        }
        else
            k = nexts[k];
    }
    /*for(int i=0;i<6;++i){
        printf("next[%d] = %d\n",i,nexts[i]);
    }*/
}
int KMP()
{
    int i=0,j=0,cnt=0;
    GetNexts();
    while(i<n){
        if(j==-1||s[i]==p[j]){
            i++;
            j++;
        }
        else
            j = nexts[j];
        if(j==m){
            cnt++;
            j = nexts[j]; //不知道为啥，没有这句话竟然也能过。。。
        }
    }
    return cnt;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
       scanf("%s%s",p,s);
       m = strlen(p);
       n = strlen(s);
       printf("%d\n",KMP());
    }
    return 0;
}

如果是让求s中有多少个单独的p，也就是p不能交叉，比如以下案例：

6(T)
abab
ababab
ans=1
abab
abababab
ans=2
aza
azazaza
ans=2

代码为（不保证正确性）：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<utility>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#define maxn1 1000005
#define maxn2 10005
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define E 1e-8
#define mod 1000000007
#define P pair<int,int>
using namespace std;

int n,m;
char s[maxn1];
char p[maxn2];
int nexts[maxn2];

void GetNexts()
{
    int j=0,k=-1;
    nexts[0] = -1;
    while(j<m){
        if(k==-1||p[j]==p[k]){
            j++;
            k++;
            nexts[j] = k;
        }
        else
            k = nexts[k];
    }
}
int KMP()
{
    int i=0,j=0,cnt=0;
    GetNexts();
    while(i<n){
        if(j==-1||s[i]==p[j]){
            i++;
            j++;
        }
        else
            j = nexts[j];
        if(j==m){
            cnt++;
            j = 0;
        }

    }
    return cnt;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
       scanf("%s%s",p,s);
       m = strlen(p);
       n = strlen(s);
       printf("%d\n",KMP());
    }
    return 0;
}