@Lizhuojin2014

咱啥也不说，刚才两个弱智贴大家都看见了吧？

还有ctj的事情，首先推理一波：

• 他黑题10左右，紫题20+。

• 他要真是大佬怎么可能才蓝名？

• 真自己写的话肯本不可能 7天内做完30题紫/黑（7天之外估值更新了，他不可能蓝名）

懂了吧？tj大佬。

我没找，目前找到一篇

P1713

#include <bits/stdc++.h>

using pii = std::pair<int,int>;
const int inf = 1e7;

int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};

int n,m,dis[12][12],ban[12][12];

std::map< std::vector<int> , int >dp[12][12];	

std::vector<int> process(std::vector<int>S) {
	int cnt = 1;
	int vis[30] = {0};
	for (int i = 1; i <= n + 1; ++ i) {
		if (S[i] == 0 || S[i] == 1) continue;
		if (!vis[S[i]]) { vis[S[i]] = ++cnt; S[i] = cnt; }
		else S[i] = vis[S[i]];
	} return S;
}
std::vector<int> shift(std::vector<int>S){ for (int i = 1; i <= n; ++ i) S[i] = S[i+1]; S[n+1] = 0; return S; }
std::vector<int> trans(std::vector<int>S,int i) { std::swap(S[i],S[i+1]); return S; }
std::vector<int> create(std::vector<int>S,int i) { S[i] = S[i+1] = 20; return process(S); }
std::vector<int> merge(std::vector<int>S,int i) {
	int p = S[i], q = S[i+1], flag = 20;
	if (p == 1 || q == 1) flag = 1;
	for (int i = 1; i <= n + 1; ++ i) if (S[i] == p || S[i] == q) S[i] = flag;
	S[i] = S[i+1] = 0;
	return process(S);
}

int main() {
	
	scanf("%d%d",&n,&m);
	for (int i = 1; i <= m; ++ i) {
		int x,y;
		scanf("%d%d",&x,&y);
		ban[x][y] = 1;
		assert( not ( (x == 1 && y == n)  or (x == n && y == 1) ) );
	}
	
	auto bfs = [&]() {
		std::memset(dis,-1,sizeof(dis));
		std::queue< pii >q;
		q.push( { 1,n } );
		dis[1][n] = 0;
		while (q.size()) {
			auto u = q.front(); q.pop();
			int x = u.first, y = u.second;
			for (int i = 0; i < 4; ++ i) {
				int x1 = x + dx[i], y1 = y + dy[i];
				if (x1 < 1 || x1 > n || y1 < 1 || y1 > n || dis[x1][y1] != -1 || ban[x1][y1]) continue;
				dis[x1][y1] = dis[x][y] + 1;
				q.push( { x1,y1 } );
			}
		} 
	};  bfs();
	
	std::vector<int>v(n+2);
	v[n+1] = 1; dp[1][n-1][v] = 1;
	v[n+1] = 0; v[n] = 1; dp[1][n-1][v] = 1;
	
	for (int i = 1; i <= n; ++ i) {
		for (int j = n - (i == 1); j >= 1; j --) {
			for (auto P:dp[i][j]) {
				auto S = P.first; int v = P.second;
				int p = S[j], q = S[j+1];
				if (p == 0 && q == 0) dp[i][j-1][S] = std::max(dp[i][j-1][S],v);
				if (ban[i][j]) continue;
				if (p == 0 && q == 0) {
					auto T = create(S,j);
					dp[i][j-1][T] = std::max(dp[i][j-1][T],v + 1);
				}
				if ((p != 0) + (q != 0) == 1) {
					dp[i][j-1][S] = std::max(dp[i][j-1][S],v + 1);
					auto T = trans(S,j);
					dp[i][j-1][T] = std::max(dp[i][j-1][T],v + 1);
				}
				if (p && q && p != q) {
					auto T = merge(S,j);
					dp[i][j-1][T] = std::max(dp[i][j-1][T],v + 1);
				}
			}
		} for (auto P:dp[i][0]) {
			auto S = P.first; int v = P.second;
			if (S[1] == 0) dp[i+1][n][shift(S)] = dp[i][0][S];
		}
	}
	std::vector<int>S1(n+2),S2(n+2);
	S1[1] = 1; S2[2] = 1;
	printf("%d",std::max(dp[n][1][S1],dp[n][1][S2]) - dis[n][1]);
	return 0;
}
他的代码 提交记录https://www.luogu.com.cn/record/list?pid=P1713&user=Lizhuojin2014&page=1

和题解区第一篇一摸一样，可以去翻翻。https://www.luogu.com.cn/problem/solution/P1713

其他的我再去找找，反正实锤了。

@chen_zhe