A - Censoring(KMP +栈）

最新推荐文章于 2023-08-06 23:08:13 发布

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#栈和KMP

本文介绍了一种文本审查算法，用于从给定的字符串S中移除所有子字符串T的实例，直至S中不再包含T。通过使用栈和KMP算法，算法能够有效地处理长度高达10^6的字符串。

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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 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

有一个S串和一个T串，长度均小于1,000,000，设当前串为U串，然后从前往后枚举S串一个字符一个字符往U串里添加，若U串后缀为T，则去掉这个后缀继续流程。

Input

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

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

题意：两个串，遍历第一个串的每个元素，每当串尾出现第二个串则删除。输出删除后的串。

题解：栈和KMP同时使用，使用栈放入一个字符，每放入一个字符做一次KMP，如果有符合串则删除，如果不符合则将该字符放入ans数组。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>

using namespace std;
const int MaxN = 1e6;

char a[MaxN + 5],b[MaxN + 5],ans[MaxN];
int len1,len2;
int kmp[MaxN + 5],q[MaxN + 5];

int main(){
	cin >> a+1;
	cin >> b+1;
	len1 = strlen(a+1);
	len2 = strlen(b+1);
	int k = 0;
	for(int i = 2;i <= len2;i++){
		while(k && b[k + 1] != b[i])k = kmp[k];
		if(b[k+1] == b[i])k++;
		kmp[i] = k;
	}
	k = 0;
	int tot = 0;
	for(int i = 1;i <= len1;i++){
		k = q[tot];
		while(k && b[k+1] != a[i])k = kmp[k];
		if(b[k+1] == a[i])k++;
		if(k == len2) tot = tot - len2 + 1;
		else{
			ans[++tot] = a[i];
			q[tot] = k;
		}
	}
	for(int i = 1;i <= tot;i++){
		cout << ans[i];
	}
	cout << endl;
}