Kattis

String Matching

Input

The input consists of several test cases. Each test case consists of two lines, first a non-empty pattern, then a non-empty text. Input is terminated by end-of-file. The input file will not be larger than 5 Mb.

Output

For each test case, output one line containing the positions of all the occurences of pattern in text, from first to last, separated by a single space.

Sample Input 1	Sample Output 1
p Popup helo Hello there! peek a boo you speek a bootiful language anas bananananaspaj	2 4 5 7

题意

多组输入，每组两行，模式串和对比串，输出上面的模式串在下面的字符串中的所有位置下标，下标从0开始

思路1

KMP算法，套个模版就可以了

思路2

用string的find，str2.find(str1,x)，关于这个函数的用法http://www.cplusplus.com/reference/string/string/find/

代码1

#include<stdio.h>
#include<string.h>
using namespace std;
int const MAXM = 4000000;
char s[MAXM], t[MAXM];
int next[MAXM], n;
int shuchu[MAXM];
void get_next()
{
    next[0] = -1;
    int i = 0, j = -1;
    while (t[i] != '')
    {
        if (j == -1 || t[i] == t[j])
        {
            i++;
            j++;
            next[i] = j;
        }
        else
            j = next[j];
    }
}

int KMP()
{
    get_next();
    int i = 0, j = 0, ans = 0, len = strlen(t);
    while (s[i])
    {
        if (j == -1 || s[i] == t[j])
        {
            i++;
            j++;
        }
        else
            j = next[j];
        if (j == len)
        {
            j = next[j];
            shuchu[ans++] = i;
        }
    }
    return ans;
}

int main()
{
    
    while (gets(t)) {
        gets(s);
        int ss = KMP();
        for (int i = 0; i < ss; i++)
        {
            if (i)printf(" ");
            printf("%d", shuchu[i]-strlen(t));
        }
        puts("");
    }
}

代码2

#include<bits/stdc++.h>
using namespace std;
int main(){
    string str1,str2;
    int cnt[100000];
    while(getline(cin,str1)){
        getline(cin,str2);
        int pos=0,x=0;
        while(str2.find(str1,x)!=-1&&x<str2.size()){
            cnt[pos++]=str2.find(str1,x);
            x=cnt[pos-1]+1;
        }
        for(int i=0;i<pos;i++){
            printf("%d%c",cnt[i],i==pos-1?'
':' ');
        }
    }
    return 0;
}