AtCoder Beginner Contest 329

比赛地址：Ｓｋｙ Inc, Programming Contest 2023（AtCoder Beginner Contest 329） - AtCoder

E - Stamp

Problem Statement

You are given two strings: $S$, which consists of uppercase English letters and has length $N$, and $T$, which also consists of uppercase English letters and has length $M(\le N)$.

There is a string $X$ of length $N$ consisting only of the character #. Determine whether it is possible to make $X$ match $S$ by performing the following operation any number of times:

• Choose $M$ consecutive characters in $X$ and replace them with $T$.

Constraints

• $1\le N\le 2\times 10^5$
• $1\le M\le \min (N,$ $5$$)$
• $S$ is a string consisting of uppercase English letters with length $N$.
• $T$ is a string consisting of uppercase English letters with length $M$.

Input

The input is given from Standard Input in the following format:

$N$ $M$
$S$
$T$

Output

Print Yes if it is possible to make $X$ match $S$; print No otherwise.

Sample Input 1

7 3
ABCBABC
ABC

Sample Output 1

Yes

Below, let $X[l:r]$ denote the part from the $l$-th through the $r$-th character of $X$.

You can make $X$ match $S$ by operating as follows.

1. Replace $X[3:5]$ with $T$. $X$ becomes ##ABC##.
2. Replace $X[1:3]$ with $T$. $X$ becomes ABCBC##.
3. Replace $X[5:7]$ with $T$. $X$ becomes ABCBABC.

题目大意

给定两个字符串$S_1,S_2$，长度分别为$n,m$，然后再给一个长度为$n$的字符串$S_3$，$S_3$是全部由‘#’组成的字符串。

题目要求每次用$S_2$来覆盖$S_3$的一段连续的位置，判断是否可以将$S_3$构造成$S_1$。

思路

我们可以将本题理解为粘胶带，我们可以将$S_3$看作是一个长度为$n$空白的板子，将$S_2$看作是一个长度为$m$且印有字母的胶带，这样每次操作就相当于在空白板子上粘胶带，而每次粘胶带时都会覆盖掉当前这段长度为$m$的位置，求最后是否能将该空白的板子粘成$S_1$的样子。

如果我们考虑根据粘胶带的顺序来判断是否能构造成$S_1$，那么就会出现有后效性，即后续操作会影响当前操作。

于是换个思路来想，根据粘胶带的思路，无论怎么粘，最后一个粘的胶带都不会被遮挡，所以我们考虑倒着操作，将$S_1$胶带一张一张揭下来，最后判断是否能揭成一个空白的板子。

因此我们枚举$S_1$中的子段$S_2$，当然其中包含形如“AB#”，这种与“ABC”等价，因为第三个位置实际上是会被覆盖的。然后用dfs一层一层的揭，在dfs搜索的时候要注意，当前位置的前面和后面都要搜索一边，因为这个字符串可能被覆盖的前缀，也可能是后缀。

实现

#include<bits/stdc++.h>

using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
#define LL long long 
#define ULL unsigned long long 
#define PII pair<int, int>
#define lowbit(x) (x & -x)
#define Mid ((l + r) >> 1)
#define ALL(x) x.begin(), x.end()
#define endl '\n'
#define fi first 
#define se second

const int INF = 0x7fffffff;
const int mod = 1e9 + 7;
const int N = 2e5 + 10; 

int n, m;
string s1, s2;

void modify(int x) {
    for(int i = x; i <= x + m - 1; i ++ ) {
        s1[i] = '#';
    }
}

bool check(int x) {
    bool flag = false;
    for(int i = 0; i < m; i ++ ) {
        if(s1[x + i] != s2[i]) {
            if(s1[x + i] != '#') 
            	return false;
        } else {
            flag = true;	//不能全为'#'
        }
    }
    return flag;
}

void dfs(int x) {
    modify(x);
    for(int i = x + 1; i < x + m && i < n; i ++ ) {
        if(check(i)) dfs(i);
    }
    for(int i = max(0, x - m + 1); i < x; i ++ ) {
        if(check(i)) dfs(i);
    }
}

void Solved() {
    cin >> n >> m;
    cin >> s1 >> s2;

    for(int i = 0; i < n - m + 1; i ++ ) {
        if(check(i)) dfs(i);
    }
    for(int i = 0; i < n; i ++ ) {
        if(s1[i] != '#') {
            cout << "No" << endl; return;
        }
    }
    cout << "Yes" << endl;
}

signed main(void) {
    IOS

    int ALL = 1; 
    // cin >> ALL;
    while(ALL -- ) Solved();

    return 0;
}