题目传送门

Oulipo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 36811 Accepted Submission(s): 13875

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

输入case数，每个case由两个字符串p，s组成，求p在s中出现的次数

思路

典型的KMP板子题，非常适合和我一样初学KMP的人。

要简单的修改一下KMP模板，P的长度为len，在求解next数组的时候，需要把next [len] 也求出来，这样在匹配到模板串的最后一位的时候，让 j = next [ j ] ，跳转到下一个匹配的位置。

举个例子： A Z A

next数组为：-1 0 -1 1

匹配的AZAZAZA的时候，首先匹配到AZAZAZA，然后$j=3=strlen(AZA)$，此是i指向Z，因为已经匹配完成，那么下一次匹配就从next[len]开始，因为如果从0开始的话，显然是错误的。因此 j 跳转到next[len]的位置。

要注意：

• 使用c++ STL string的时候，代码中不要用int值和string.size()比较大小，string.size()返回的是一个unsigned int，为无符号整数，$(int)-1>(unsignedint)1$，因为编码方式不同，$(unsignedint)－1=2^{32}-1=4294967295$

另外我本来想的是不改KMP算法，假如在s中找到p的位置为pos，就把s[pos]的字符改掉，重新走一遍KMP…然后就T掉了…初学还是理解的不够透彻啊QAQ

关于KMP算法的讲解请看大牛博客：KMP算法详解-彻底清楚了

之后有时间会写我个人的理解~

代码

#include <bits/stdc++.h>
#define MAXN 1000010
using namespace std;
int nex[MAXN];
char a[MAXN];
char b[MAXN];
void getNext(char *p){int j=0;int k=-1;nex[0]=-1;int len =strlen(p);while (j<len){if (k==-1 || p[j]==p[k]){if (p[++j]==p[++k]){nex[j]=nex[k];}else {nex[j]=k;}}else {k=nex[k];}}
}
int kmp(char *s,char *p){int len=strlen(p);int len2=strlen(s);int i=0,j=0;int ans=0;while (i< len2 && j < len){if (j==-1 || s[i]==p[j]){i++;j++;}else {j=nex[j];}if (j==len){ans++;j=nex[j];}}return ans;
}int main (){int n;scanf("%d",&n);while (n--){scanf("%s%s",&a,&b);int ans=0;getNext(a);printf("%d\n",kmp(b,a));
//      原始写法（ 会T ）
//      int pos=kmp(b,a);
//        while (pos!=-1){
//            ans++;
//            b[pos]='?';
//            pos=kmp(b,a);
//        }
//      printf("%d\n",pos);}return 0;
}

KMP模板

注释掉的是C版本

#include <string>
#include <iostream>
#define MAXN 1000010
using namespace std;
int nex[MAXN];
//char a[MAXN];
//char b[MAXN];
void getNext(string p){int len=p.size();
//void getNext(char *p){
//    int len =strlen(p);int j=0;int k=-1;nex[0]=-1;while (j<len-1){//这里也可以写 len，看自己的情况，本题就改成了lenif (k==-1 || p[j]==p[k]){if (p[++j]==p[++k]){nex[j]=nex[k];}else {nex[j]=k;}}else {k=nex[k];}}}
//int kmp(char *s,char *p){
//    int len=strlen(p);
//    int len2=strlen(s);
int kmp(string s,string p){
int len=p.size();
int len2=s.size();int i=0,j=0;int ans=0;while (i< len2 && j < len){if (j==-1 || s[i]==p[j]){i++;j++;}else {j=nex[j];}}if (j>=len){return i-j+1;}else {return -1;}
}int main (){
//    std::ios::sync_with_stdio(false);int n;string a,b;
//    while (scanf("%s%s",&a,&b)!=EOF){while (cin>>a>>b){getNext(a);cout<<kmp(b,a)<<endl;
//      printf("%d\n",kmp(b,a));}return 0;
}