CF1178F1 Short Colorful Strip 题解

Short Colorful Strip

传送门

题面翻译

题目描述

这是F题的第一个子任务。F1和F2的区别仅在对于m和时间的限制上

有n+1种颜色标号从0到n，我们有一条全部染成颜色0的长为m的纸带。

Alice拿着刷子通过以下的过程来给纸带染色：

我们按照从1到n的顺序进行染色，进行每次染色时，我们选取一个区间[ai,bi]，0<=ai<bi<=m，并且这个区间内必定是单种颜色。

染色到最后，纸带上有各种颜色除了颜色0.给出纸带最终的状态，问有多少种不同的染色方案能到达最终状态。输出时结果模998244353。

输入格式

第一行有两个整数n和m，1<=n<=500，并且保证m=n。n代表除了颜色0有多少种颜色，m代表纸带的长度。

第二行有m个用空格分隔的整数c1,c2,…,cm,(1<=ci<=n)，代表完成染色后纸带的最终状态。保证最终状态中能在纸带上找到从1到n所有的颜色。

注意，这个子任务保证了n=m，那么就是说最终状态是1到n的一个排列。

输出格式

输入一个单独的整数，即Alice可行的染色方案数模998244353.

题目描述

This is the first subtask of problem F. The only differences between this and the second subtask are the constraints on the value of $m$ and the time limit. You need to solve both subtasks in order to hack this one.

There are $n + 1$ distinct colours in the universe, numbered $0$ through $n$ . There is a strip of paper $m$ centimetres long initially painted with colour $0$ .

Alice took a brush and painted the strip using the following process. For each $i$ from $1$ to $n$ , in this order, she picks two integers $\leq a_i < b_i \leq m$ , such that the segment $a_i, b_i]$ is currently painted with a single colour, and repaints it with colour $i$ .

Alice chose the segments in such a way that each centimetre is now painted in some colour other than $0$ . Formally, the segment $[i - 1, i]$ is painted with colour $c_i$ ( $c_i \neq 0$ ). Every colour other than $0$ is visible on the strip.

Count the number of different pairs of sequences ${a_i\}_{i=1}^n$ , ${b_i\}_{i=1}^n$ that result in this configuration.

Since this number may be large, output it modulo $998244353$ .

输入格式

The first line contains a two integers $n$ , $m$ ( $\leq n \leq 500$ , $n = m$ ) — the number of colours excluding the colour $ 0 $ and the length of the paper, respectively.

The second line contains $m$ space separated integers $c_1, c_2, \ldots, c_m$ ( $\leq c_i \leq n$ ) — the colour visible on the segment $[i - 1, i]$ after the process ends. It is guaranteed that for all $j$ between $1$ and $n$ there is an index $k$ such that $c_k = j$ .

Note that since in this subtask $n = m$ , this means that $c$ is a permutation of integers $1$ through $n$ .

输出格式

Output a single integer — the number of ways Alice can perform the painting, modulo $ 998244353 $ .

样例 #1

样例输入 #1

3 3
1 2 3

样例输出 #1

样例 #2

样例输入 #2

7 7
4 5 1 6 2 3 7

样例输出 #2

提示

In the first example, there are $5$ ways, all depicted in the figure below. Here, $0$ is white, $1$ is red, $2$ is green and $3$ is blue.

Below is an example of a painting process that is not valid, as in the second step the segment 1 3 is not single colour, and thus may not be repainted with colour $2$ .

解题思路

有题目描述可知， $n = m$ ，所以每个颜色只能涂一次。又可知，颜色要从编号小的到编号大的涂，所以可以用搜索算法，每次找出带子上颜色编号最小的位置作为基准点，算出 $df s (l, mini d - 1)$ 与 $df s (mini d + 1, r)$ 。而题目要求求涂色方案数，所以可以用乘法原理，算出 $\times dfs(x,minid-1)$ 、 $\times dfs(x+1,r)$ ，因而 $\times dfs(midid+1,r)=dfs(minid,x) \times dfs(x+1,r)$ 。

为了避免做重复涂色操作，所以要记忆化，用 $tong_{l,r}$
记录 $l$ ~ $r$ 区间的涂色方案数。

AC Code

// C++ includes used for precompiling -*- C++ -*-// Copyright (C) 2003-2013 Free Software Foundation, Inc.
//
// This file is part of the GNU ISO C++ Library.  This library is free
// software; you can redistribute it and/or modify it under the
// terms of the GNU General Public License as published by the
// Free Software Foundation; either version 3, or (at your option)
// any later version.// This library is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.// Under Section 7 of GPL version 3, you are granted additional
// permissions described in the GCC Runtime Library Exception, version
// 3.1, as published by the Free Software Foundation.// You should have received a copy of the GNU General Public License and
// a copy of the GCC Runtime Library Exception along with this program;
// see the files COPYING3 and COPYING.RUNTIME respectively.  If not, see
// <Licenses - GNU Project - Free Software Foundation>./** @file stdc++.h*  This is an implementation file for a precompiled header.*/// 17.4.1.2 Headers// C
#ifndef _GLIBCXX_NO_ASSERT#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>#if __cplusplus >= 201103L#include <ccomplex>#include <cfenv>#include <cinttypes>#include <cstdalign>#include <cstdbool>#include <cstdint>#include <ctgmath>#include <cwchar>#include <cwctype>
#endif// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>#if __cplusplus >= 201103L#include <array>#include <atomic>#include <chrono>#include <condition_variable>#include <forward_list>#include <future>#include <initializer_list>#include <mutex>#include <random>#include <ratio>#include <regex>#include <scoped_allocator>#include <system_error>#include <thread>#include <tuple>#include <typeindex>#include <type_traits>#include <unordered_map>#include <unordered_set>
#endif
using namespace std;
#define int long long
const int Mod = 998244353;
const int Maxn = 500 + 5;
int n, m, a[Maxn];
int tong[Maxn][Maxn];//区块l~r的方案数
inline int dfs(int l, int r) {if (tong[l][r]) {return tong[l][r];} else if (l >= r) {tong[l][r] = 1;} else {int minid = l, tmp1 = 0, tmp2 = 0;for (int i = l; i <= r; i++) {//求当前区块编号最小颜色if (a[i] < a[minid]) {minid = i;}}for (int i = l; i <= minid; i++) {tmp1 = (tmp1 + dfs(l, i - 1) * dfs(i, minid - 1) % Mod) % Mod;}for (int i = minid; i <= r; i++) {tmp2 = (tmp2 + dfs(minid + 1, i) * dfs(i + 1, r) % Mod) % Mod;}tong[l][r] = tmp1 * tmp2 % Mod;}return tong[l][r];
}
inline void work() {cin >> n >> m;for (int i = 1; i <= n; i++) {cin >> a[i];}dfs(1, m);cout << tong[1][m] % Mod << endl;
}
signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);work();return 0;
}