嘟嘟嘟

都说这题是斯坦纳树的板儿题。

斯坦纳树，我也不知道为啥起这么个名儿，斯坦纳树主要用来解决这样一类问题：带边权无向图上有几个（一般约10个）点是【关键点】，要求选择一些边使这些点在同一个联通块内，同时要求所选的边的边权和最小。（摘自兔哥博客）
但说白了就是一种状压dp。令\(dp[i][j][S]\)表示和点\((i, j)\)相连的关键点的状态为\(S\)时的最小代价，于是有两个转移方程：
\[\begin{align*} dp[i][j][S] &= min_{k \in S} \{ dp[i][j][k] + dp[i][j][\complement_kS] - a[i][j] \} \\ dp[i][j][S] &= min \{ dp[x][y][S] + a[i][j] \} \end{align*}\]
第一个方程很好转移；第二个方程必须满足\((i, j)\)和\((x, y)\)相邻，这个用刷表法就不行了，但是可以用类似spfa的方法转移！

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 12;
inline ll read()
{ll ans = 0;char ch = getchar(), last = ' ';while(!isdigit(ch)) last = ch, ch = getchar();while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();if(last == '-') ans = -ans;return ans;
}
inline void write(ll x)
{if(x < 0) x = -x, putchar('-');if(x >= 10) write(x / 10);putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclrfreopen("ha.in", "r", stdin);freopen("ha.out", "w", stdout);
#endif
}int n, m, N, a[maxn][maxn], cnt = 0;struct Node {int x, y, S;} pre[maxn][maxn][1 << maxn];
int dp[maxn][maxn][1 << maxn];#define pr pair<int, int>
#define mp make_pair
#define fir first
#define sec second
const int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
bool in[maxn][maxn];
queue<pr> q;
In void spfa(int S)
{while(!q.empty()){int x = q.front().fir, y = q.front().sec;q.pop(); in[x][y] = 0;for(int i = 0; i < 4; ++i){int nx = x + dx[i], ny = y + dy[i];if(nx <= 0 || nx > n || ny <= 0 || ny > m) continue;if(dp[nx][ny][S] > dp[x][y][S] + a[nx][ny]){dp[nx][ny][S] = dp[x][y][S] + a[nx][ny];pre[nx][ny][S] = (Node){x, y, S};if(!in[nx][ny]) q.push(mp(nx, ny)), in[nx][ny] = 1;}}}
}bool vis[maxn][maxn];
In void dfs(int x, int y, int S)
{vis[x][y] = 1; Node tp = pre[x][y][S];if(!tp.x) return;dfs(tp.x, tp.y, tp.S);if(tp.x == x && tp.y == y) dfs(tp.x, tp.y, S ^ tp.S);
}int main()
{
//  MYFILE();n = read(), m = read();Mem(dp, 0x3f);for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j){a[i][j] = read();if(!a[i][j]) dp[i][j][1 << (cnt++)] = 0;}N = 1 << cnt;for(int S = 0; S < N; ++S){for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j){for(int k = S, tp; k; k = S & (k - 1))if(dp[i][j][S] > (tp = dp[i][j][S ^ k] + dp[i][j][k] - a[i][j])){dp[i][j][S] = tp;pre[i][j][S] = (Node){i, j, k};}if(dp[i][j][S] ^ INF) q.push(mp(i, j)), in[i][j] = 1;}spfa(S);}int x, y, ans = INF;for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j) if(dp[i][j][N - 1] < ans) ans = dp[i][j][N - 1], x = i, y = j;write(ans), enter;dfs(x, y, (1 << cnt) - 1);for(int i = 1; i <= n; ++i, enter)for(int j = 1; j <= m; ++j){if(!a[i][j]) putchar('x');else putchar(vis[i][j] ? 'o' : '_');}return 0;
}