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String Matching

[编程题]String Matching

热度指数：11045 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 64M，其他语言128M
算法知识视频讲解

Finding all occurrences of a pattern in a text is a problem that arises frequently in text-editing programs. Typically,the text is a document being edited,and the pattern searched for is a particular word supplied by the user. We assume that the text is an array T[1..n] of length n and that the pattern is an array P[1..m] of length m<=n.We further assume that the elements of P and T are all alphabets(∑={a,b...,z}).The character arrays P and T are often called strings of characters. We say that pattern P occurs with shift s in the text T if 0<=s<=n and T[s+1..s+m] = P[1..m](that is if T[s+j]=P[j],for 1<=j<=m). If P occurs with shift s in T,then we call s a valid shift;otherwise,we calls a invalid shift. Your task is to calculate the number of vald shifts for the given text T and p attern P.

输入描述:

   For each case, there are two strings T and P on a line,separated by a single space.You may assume both the length of T and P will not exceed 10^6.

输出描述:

    You should output a number on a separate line,which indicates the number of valid shifts for the given text T and pattern P.

示例1

输入

abababab abab

输出

算法知识视频讲解

胃溃疡

#include <stdio.h>

#include <string.h>

#define N 1000001

int main()

{

char str1[N],str2[N];

scanf("%s%s",str1,str2);

char*a=str1;

int len1=strlen(str1);

int len2=strlen(str2);

int count=0;

a=strstr(str1,str2);

while (a!=NULL) {

count++;

a=a+1;

a=strstr(a,str2);

}

printf("%d\n",count);

}

发表于 2023-03-02 10:37:06 回复(0)

稀有的猪

使用kmp算法

#include <string.h>
#include <stdio.h>
#include <stdlib.h>

#define maxn 1000010

char s[maxn];
char t[maxn];


void build_match(char *t, int *match)
{
    match[0] = 0;
    size_t i = 0, j = 1;
    while (j < strlen(t))
    {
        if (t[i] == t[j])
        {
            match[j] = i + 1;
            i++, j++;
        }
        else 
        {
            if (i != 0)    
                i = match[i - 1];       
            else            
            {
                match[j] = 0;
                j++;
            }          
        }
    }
}

int kmp(char *s, char *t)
{
    int i = 0, j = 0, n = strlen(s), m = strlen(t);
    int sum = 0;
    int *match = (int*)malloc(sizeof(int) * m);
    build_match(t, match);
    while (i < n)
    {
        if (s[i] == t[j])
        {
            i++;
            j++;
        }
        else 
        {
            if (j != 0)
            {
                j = match[j - 1];
            }
            else
            {
                i++;
            }
        }       

        if (j == m)
        {
            sum++;
            j = match[j - 1];
        }
    }
    free(match);
    return sum;
}

int main()
{
    while (scanf("%s %s",s, t) != EOF)
    {
        printf("%d\n", kmp(s, t));
    }
    return 0;
}

朴素搜索算法

#include <stdio.h>
#include <string.h>

#define maxn 1000010

char s[maxn];
char t[maxn];

int match(char *s, char *t)
{
    size_t i = 0, j = 0, k = 0;
    int sum = 0;
    while (i < strlen(s))
    {
        if (s[i] == t[j])
        {
            i++;j++;
        }
        else
        {
            j = 0;
            k++;
            i = k; 
        }
        if (j == strlen(t))
        {
            sum++;
            j = 0;
            k++;
            i = k;
        }
    }
    return sum;
}

int main()
{
    while(scanf("%s %s",s, t) != EOF)
    {
        printf("%d\n", match(s, t));
    }
    return 0;
}

发表于 2022-03-07 13:29:06 回复(0)