【CF1234F】 Yet Another Substring Reverse

题目

题目描述
Let’s define p_i(n)p
i
​
(n) as the following permutation: [i, 1, 2, \dots, i - 1, i + 1, \dots, n][i,1,2,…,i−1,i+1,…,n] . This means that the ii -th permutation is almost identity (i.e. which maps every element to itself) permutation but the element ii is on the first position. Examples:

p_1(4) = [1, 2, 3, 4]p
1
​
(4)=[1,2,3,4] ;
p_2(4) = [2, 1, 3, 4]p
2
​
(4)=[2,1,3,4] ;
p_3(4) = [3, 1, 2, 4]p
3
​
(4)=[3,1,2,4] ;
p_4(4) = [4, 1, 2, 3]p
4
​
(4)=[4,1,2,3] .
You are given an array x_1, x_2, \dots, x_mx
1
​
,x
2
​
,…,x
m
​
( 1 \le x_i \le n1≤x
i
​
≤n ).

Let pos(p, val)pos(p,val) be the position of the element valval in pp . So, pos(p_1(4), 3) = 3, pos(p_2(4), 2) = 1, pos(p_4(4), 4) = 1pos(p
1
​
(4),3)=3,pos(p
2
​
(4),2)=1,pos(p
4
​
(4),4)=1 .

Let’s define a function f§ = \sum\limits_{i=1}^{m - 1} |pos(p, x_i) - pos(p, x_{i + 1})|f§=
i=1
∑
m−1
​
∣pos(p,x
i
​
)−pos(p,x
i+1
​
)∣ , where |val|∣val∣ is the absolute value of valval . This function means the sum of distances between adjacent elements of xx in pp .

Your task is to calculate f(p_1(n)), f(p_2(n)), \dots, f(p_n(n))f(p
1
​
(n)),f(p
2
​
(n)),…,f(p
n
​
(n)) .

输入格式
The first line of the input contains two integers nn and mm ( 2 \le n, m \le 2 \cdot 10^52≤n,m≤2⋅10
5
) — the number of elements in each permutation and the number of elements in xx .

The second line of the input contains mm integers ( mm , not nn ) x_1, x_2, \dots, x_mx
1
​
,x
2
​
,…,x
m
​
( 1 \le x_i \le n1≤x
i
​
≤n ), where x_ix
i
​
is the ii -th element of xx . Elements of xx can repeat and appear in arbitrary order.

输出格式
Print nn integers: f(p_1(n)), f(p_2(n)), \dots, f(p_n(n))f(p
1
​
(n)),f(p
2
​
(n)),…,f(p
n
​
(n)) .

题意翻译
给定一个长度为mm的序列xx

记：

排列p_i(n)=[i,1,2,…,i-1,i+1,…,n]p
i
​
(n)=[i,1,2,…,i−1,i+1,…,n] pos(p,x)pos(p,x)表示xx在排列pp中的第几位 f§=\sum_{i=1}^{m-1}|pos(p,x_i)-pos(p,x_{i+1})|f§=∑
i=1
m−1
​
∣pos(p,x
i
​
)−pos(p,x
i+1
​
)∣

求f(p_1(n)),f(p_2(n)),…,f(p_n(n))f(p
1
​
(n)),f(p
2
​
(n)),…,f(p
n
​
(n))。

2\le n,m\le2\times10^5,1\le x_i\le n2≤n,m≤2×10
5
,1≤x
i
​
≤n

输入输出样例
输入 #1复制
4 4
1 2 3 4
输出 #1复制
3 4 6 5
输入 #2复制
5 5
2 1 5 3 5
输出 #2复制
9 8 12 6 8
输入 #3复制
2 10
1 2 1 1 2 2 2 2 2 2
输出 #3复制
3 3
说明/提示
Consider the first example:

x = [1, 2, 3, 4]x=[1,2,3,4] , so

for the permutation p_1(4) = [1, 2, 3, 4]p
1
​
(4)=[1,2,3,4] the answer is |1 - 2| + |2 - 3| + |3 - 4| = 3∣1−2∣+∣2−3∣+∣3−4∣=3 ;
for the permutation p_2(4) = [2, 1, 3, 4]p
2
​
(4)=[2,1,3,4] the answer is |2 - 1| + |1 - 3| + |3 - 4| = 4∣2−1∣+∣1−3∣+∣3−4∣=4 ;
for the permutation p_3(4) = [3, 1, 2, 4]p
3
​
(4)=[3,1,2,4] the answer is |2 - 3| + |3 - 1| + |1 - 4| = 6∣2−3∣+∣3−1∣+∣1−4∣=6 ;
for the permutation p_4(4) = [4, 1, 2, 3]p
4
​
(4)=[4,1,2,3] the answer is |2 - 3| + |3 - 4| + |4 - 1| = 5∣2−3∣+∣3−4∣+∣4−1∣=5 .
Consider the second example:

x = [2, 1, 5, 3, 5]x=[2,1,5,3,5] , so

for the permutation p_1(5) = [1, 2, 3, 4, 5]p
1
​
(5)=[1,2,3,4,5] the answer is |2 - 1| + |1 - 5| + |5 - 3| + |3 - 5| = 9∣2−1∣+∣1−5∣+∣5−3∣+∣3−5∣=9 ;
for the permutation p_2(5) = [2, 1, 3, 4, 5]p
2
​
(5)=[2,1,3,4,5] the answer is |1 - 2| + |2 - 5| + |5 - 3| + |3 - 5| = 8∣1−2∣+∣2−5∣+∣5−3∣+∣3−5∣=8 ;
for the permutation p_3(5) = [3, 1, 2, 4, 5]p
3
​
(5)=[3,1,2,4,5] the answer is |3 - 2| + |2 - 5| + |5 - 1| + |1 - 5| = 12∣3−2∣+∣2−5∣+∣5−1∣+∣1−5∣=12 ;
for the permutation p_4(5) = [4, 1, 2, 3, 5]p
4
​
(5)=[4,1,2,3,5] the answer is |3 - 2| + |2 - 5| + |5 - 4| + |4 - 5| = 6∣3−2∣+∣2−5∣+∣5−4∣+∣4−5∣=6 ;
for the permutation p_5(5) = [5, 1, 2, 3, 4]p
5
​
(5)=[5,1,2,3,4] the answer is |3 - 2| + |2 - 1| + |1 - 4| + |4 - 1| = 8∣3−2∣+∣2−1∣+∣1−4∣+∣4−1∣=8 .

思路

可以发现通过翻转操作我们可以将任意两个子串拼在一起。
这样子串所在的位置就不重要了，只需要记录字符的状态即可。
注意到 $|$ 字符集 $|\le 20$ ，考虑状压记录所有的字符集状态。
设 $f_s$ 表示状态 $s$ 中有多少字母（其实就是 $s$ 的二进制中有多少 $1$ ），设全集为 $T$ 。
那么：
$$f_s=\underset{1\le i\le j\le |S|}{\max }(j-i+1|S_i\ne S_{i+1}\ne ...\ne S_j,s=S_i+...+S_j)$$
所以：
$$Ans=\underset{s\subseteq T}{\max }(f_s+\underset{t\subseteq T-s}{\max }{f}_t)$$
可以发现后面的枚举子集完全可以预处理出来。
记 $g_s={\max }_{t\subseteq s}{f}_t$ ，则 $Ans={\max }_{s\subseteq T}(f_s+g_{T-s})$
总时间复杂度 $O(|S||T|+|T|2^{|T|})$

代码

#include 
 
using namespace std;
 
const int N = 20;
 
const int M = 10e6 + 10;
 
int dp[1 << N];
 
int main() {
	char s[M];
	scanf("%s", s);
	int len = strlen(s);
	for (int i = 0; i < len; i ++ ) {
		int now = 0;
		for (int j = 0; j < N && i + j < len; j ++ ) {
			int ch = s[i + j] - 'a';
			if (now & (1 << ch)) {
				break ;
			}
			now |= (1 << ch);
			dp[now] = j + 1;
		}
	}
	for (int now = 0; now < (1 << N); now ++ ) {
		for (int i = 0; i < N; i ++ ) {
			if (now & (1 << i)) {
				dp[now] = max(dp[now], dp[now ^ (1 << i)]);
			}
		}
	}
	int ans = 0;
	for (int now = 0; now < (1 << N); now ++ ) {
		int tmp = ((1 << N) - 1) ^ now;
		ans = max(ans, dp[now] + dp[tmp]);
	}
	printf("%d\n", ans);
	return 0;
}