[USACO2015February,Bronze] Problem1.Censoring(Bronze)

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本文介绍了一个简单的算法，用于从一个较长的字符串中审查并删除所有指定子字符串的出现，直至原字符串不再包含该子字符串。通过逐步加入字符并检查是否形成目标子字符串的方式实现审查过程。

Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty of material to read while waiting around in the barn during milking sessions. Unfortunately, the latest issue contains a rather inappropriate article on how to cook the perfect steak, which FJ would rather his cows not see (clearly, the magazine is in need of better editorial oversight).

FJ has taken all of the text from the magazine to create the string S of length at most 10^6 characters. From this, he would like to remove occurrences of a substring T of length <= 100 characters to censor the inappropriate content. To do this, Farmer John finds the first occurrence of T in S and deletes it. He then repeats the process again, deleting the first occurrence of T again, continuing until there are no more occurrences of T in S. Note that the deletion of one occurrence might create a new occurrence of T that didn’t exist before.

Please help FJ determine the final contents of S after censoring is complete.

INPUT FORMAT: (file censor.in)

The first line will contain S. The second line will contain T. The length of T will be at most that of S, and all characters of S and T will be lower-case alphabet characters (in the range a..z).

OUTPUT FORMAT: (file censor.out)

The string S after all deletions are complete. It is guaranteed that S will not become empty during the deletion process.

SAMPLE INPUT:

whatthemomooofun
moo

SAMPLE OUTPUT:

whatthefun

[Problem credits: Mark Gordon, 2015]

每次加入一个字符,当这个字符和前面的字符构成子串T则删除之,代码非常简短,时间复杂度是 $O(TS)$ .

我的代码:

#include<cstdio>
#include<string>
#include<iostream>
using namespace std;
string S, T, R;
int main()
{
  freopen("censor.in", "r", stdin);
  freopen("censor.out", "w", stdout);
  cin >> S >> T;
  for(int i = 0; i < S.size(); i++)
  {
    R += S[i];
    if(R.size() >= T.size() && R.substr(R.size() - T.size()) == T)
      R.resize(R.size() - T.size());
  }
  cout << R << endl;
  return 0;
}