[Question Solution] LuoGu3943: Starry Sky

The original title portal
because $k<=8$ so think of

The main idea is to
make the light dim to 1, and the light to 0. For
example, the case of a row of lights is
$a:01001100$
Then the corresponding difference array is
$b:011010100$
可以发现$a_i=b_{1}xor{b}_{2}xorr...xor{b}_i$
Then for the interval $[l,r]$ performs the inversion operation, the performance in the difference array is
$b_l=b_{l}xor1,b_{r+1}=b_{r+1}xor1$

The problem can be transformed into $b$ array becomes all 0
findbbThe number of$b$ array 1 must be an even number, and at most$2K$ th
conceivable to eliminate every two 1, to eliminate the cost of both a pre-determined by the next state is pressed

If you want to put $10000001$ becomes all 0s, but I can only flip the length to 3, 4
can first$00010001$ means to$[1,3]$ Flip
again$00000000$ means to$[4,7]$ Flip to
indicate that the difference between the two is$x$ of$1.\; To$ change them to 0, choose$x=b_1+b_2+...+b_{and}$
Multiple flip combinations are possible

Or put $00110000$ becomes all 0s, but I can only flip the length to 3, 4
can first$00010010$ means to$[3,6]$ Flip
again$00000000$ means to$[4,6]$ Reversal
shows that subtraction between reversal interval lengths is also possible

So for a certain length, pass bbYou can get your own length by adding, adding, subtracting, and subtracting the$b$ array. Find the minimum number of flips in a reasonable situation, and get it through a complete backpack

Code：

#include <bits/stdc++.h>
#define maxn 40010
#define maxm 100010
using namespace std;
int a[maxn], dp[maxm], cost[maxn], b[maxn], pos[maxn], power[25], tot, n, m, k;

inline int read(){
    
    
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

int main(){
    
    
	n = read(), k = read(), m = read();
	for (int i = 1; i <= k; ++i){
    
    
		int x = read();
		a[x] ^= 1, a[x + 1] ^= 1;
	}
	for (int i = 1; i <= n; ++i) cost[i] = 1e9;
	for (int i = 1; i <= m; ++i) b[i] = read();
	for (int i = 1; i <= m; ++i)
		for (int j = b[i]; j <= n; ++j) cost[j] = min(cost[j], cost[j - b[i]] + 1);
	for (int i = 1; i <= m; ++i)
		for (int j = n - b[i]; j >= 0; --j) cost[j] = min(cost[j], cost[j + b[i]] + 1);
	for (int i = 1; i <= n + 1; ++i)
		if (a[i]) pos[++tot] = i;
	power[0] = 1;
	for (int i = 1; i <= 20; ++i) power[i] = power[i - 1] << 1;
	for (int i = 1; i < power[tot]; ++i) dp[i] = 1e9;
	for (int i = 0; i < power[tot]; ++i)
		for (int j = 1; j <= tot; ++j)
			if (!(i & power[j - 1]))
				for (int k = j + 1; k <= tot; ++k)
					if (!(i & power[k - 1]))
						dp[i | power[j - 1] | power[k - 1]] = min(dp[i | power[j - 1] | power[k - 1]], dp[i] + cost[pos[k] - pos[j]]);
	printf("%d\n", dp[power[tot] - 1]);
	return 0;
}