HDU1686Oulipo

最新推荐文章于 2021-12-11 18:48:23 发布

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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): 5535 Accepted Submission(s): 2211

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

Source

华东区大学生程序设计邀请赛_热身赛

Recommend

lcy | We have carefully selected several similar problems for you: 1358 1711 3746 2203 3068

题目大意就是求子串在主串中出现的次数

这么熟悉。。原来在POJ做过。。

主要是NEXT数组的应用。NEXT数组是反应的是子串的前缀和后缀相同的长度。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>

#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b){
    return a>b;
}
char t[1000005];
char p[10005];
int f[10005];
void getFail(char *p,int *f)
{
    int i,j;
    f[0]=0;
    f[1]=0;
    int m=strlen(p);
        for(int i=1;i<m;i++)
        {
            j=f[i];
            while(j&&p[i]!=p[j])
            {
                j=f[j];
            }
            f[i+1]=p[i]==p[j]?j+1:0;
        }
}
int find(char * t,char *p,int *f)
{
    int sum=0;
    int n=strlen(t),m=strlen(p);
    getFail(p,f);
    int j=0;
    for(int i=0;i<n;i++)
    {
        while(j&&p[j]!=t[i])
        {
            j=f[j];
        }
        if(p[j]==t[i])
            j++;
        if(j==m)
        {
            sum++;
            j=f[j];///回溯找到上一个失配点（对称性）
        }
    }
    return sum;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    cin>>n;
    while(n--)
    {
        scanf("%s",&p);
        scanf("%s",&t);
        cout<<find(t,p,f)<<endl;
    }
    return 0;
}