原始题目

Oulipo

• Time Limit: 3000/1000 MS (Java/Others)
  
• Memory Limit: 32768/32768 K (Java/Others)
• Total Submission(s): 21478
  
• Accepted Submission(s): 8336

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

Source

华东区大学生程序设计邀请赛_热身赛

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题目大意

• 计算T中W的出现次数，W的所有连续字符必须与T的连续字符完全匹配，可能重叠。

解题思路

• 裸KMP

解题代码

//#include <bits/stdc++.h>
#include <cstdio> 
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <iomanip>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <queue>
#include <vector>
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
#define PI acos((double)-1)
#define E exp(double(1))
#define K 1000000+9
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define ms(x,a) memset((x),(a),sizeof(a))
int nt[10006];
char a[K],b[10005];
int n,m;
//参数为模板串和next数组
//字符串均从下标0开始
void kmpGetNext(char *s,int *Next)
{
    Next[0]=0;
    int len=strlen(s);
    for(int i=1,j=0;i<len;i++)
    {
        while(j&&s[i]!=s[j]) j=Next[j];
        if(s[i]==s[j]) j++;
        Next[i+1]=j;
    }
//    Next[len]=0;
}
int kmp(char *ss,char *s,int *Next)
{
    kmpGetNext(s,Next);
//  调试输出Next数组
//	int len=strlen(s);
//	for(int i=0;i<=len;i++)
//		cout<<Next[i]<<" ";
//	cout<<endl; 

    int ans=0;
    int len1=strlen(ss);
    int len2=strlen(s);
    for(int i=0,j=0;i<len1;i++)
    {
        while(j&&ss[i]!=s[j])j=Next[j];
        if(ss[i]==s[j]) j++;
        if(j==len2){
        	ans++;
//			return i-len2+2;
		}
           
    }
//    return -1;
	return ans;
}

int main(void)
{
    int t;
    scanf("%d",&t); 
    while(t--)
    {
    	memset(a,0,sizeof(a));
    	memset(b,0,sizeof(b));
    	scanf("%s",b);
    	scanf("%s",a);
    	cout<< kmp(a,b,nt)<<endl;
    }

    return 0;
}

收获与反思

• KMP水题