题目
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define fi first
#define se second
const int maxn = 2e5 + 5;
const int inf = 1e9 + 5;using namespace std;int n, m;
//int pre[maxn];
int a[30][30], b[30][30];int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};bool hascycle;
bool vis[30][30];
int ind[30][30];void dfs(int x, int y, int px, int py){//dfs找无向图的环：记录前驱结点，若后继结点已经访问过且不是前驱结点，说明找到了环if(hascycle) return;
// if(x < 0 || x > n - 1 || y < 0 || y > m - 1 || a[x][y] == 0) return;for(int i = 0; i < 4; i++){int mx = x + dx[i], my = y + dy[i];if(mx < 0 || mx > n - 1 || my < 0 || my > m - 1 || a[mx][my] == 0) continue;//是mx不是x！！！！！！！// if(x < 0 || x > n - 1 || y < 0 || y > m - 1 || a[x][y] == 0) continue;if(vis[mx][my]){if(mx == px && my == py) continue;else{hascycle = 1;return;}}vis[mx][my] = 1;dfs(mx, my, x, y);}
}// bool hascycle(){//拓扑排序找环：入度小于等于1的结点入队，遍历相邻结点，使其入度减1，若减完入度小于等于1，则入队。//得用vis判断结点是否已经入过队，不然会死循环！！！！！！！！！！！！！！
// queue<pair<int, int>> q;
// int cnt = 0;
// bool vis[30][30] = {0};
// for(int i = 0; i < n; i++){
// for(int j = 0; j < m; j++){
// if(a[i][j] == 1 && ind[i][j] <= 1){
// q.push({i, j});
// cnt++;
// vis[i][j] = 1;
// }
// }
// }
// while(!q.empty()){
// auto u = q.front();
// q.pop();
// int x = u.first, y = u.second;
// for(int i = 0; i < 4; i++){
// int mx = x + dx[i], my = y + dy[i];
// if(mx < 0 || mx > n - 1 || my < 0 || my > m - 1 || a[mx][my] == 0) continue;
// ind[mx][my]--;
// if(ind[mx][my] <= 1 && !vis[mx][my]){
// q.push({mx, my});
// vis[mx][my] = 1;
// cnt++;
// }
// }
// }
// int tot = 0;
// for(int i = 0; i < n; i++){
// for(int j = 0; j < m; j++){
// if(a[i][j]) tot++;
// }
// }
// return cnt != tot;//是不等于！！！！！
// }int calc(){hascycle = 0;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){vis[i][j] = 0;ind[i][j] = 0;}}bool flag = 0;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(a[i][j] == 1 && !flag){vis[i][j] = 1;dfs(i, j, -1, -1);flag = 1;}}}for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(a[i][j] == 1 && !vis[i][j]) return 0;}}
// for(int i = 0; i < n; i++){
// for(int j = 0; j < m; j++){
// if(a[i][j] == 0) continue;
// for(int k = 0; k < 4; k++){
// int mx = i + dx[k];
// int my = j + dy[k];
// if(mx < 0 || mx > n - 1 || my < 0 || my > m - 1 || a[mx][my] == 0) continue;
// ind[mx][my]++;
// }
// }
// }if(hascycle) return 0; int res = 0;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(a[i][j] == 1){res++;}}}return res;
}void solve(){int q;cin >> n >> m;vector<int> vec(n);int res = 0;bool flag = 0;if(n > m){swap(n, m);flag = 1;}if(n == 1){if(!flag) cout << string(m, '.') << '\n';else{for(int i = 1; i <= m; i++){cout << ".\n";}}return;}string ans[30];if(n * m <= 20){for(int x = 0; x < (1LL << (n * m)); x++){// for(int x = 0; x <= 3; x++){// int x = 3;// xx = x;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){a[i][j] = x >> (i * m + j) & 1;}}hascycle = 0;int t = calc();// cout << x << ' ' << t << ' ' << hascycle << '\n';if(t > res){for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){b[i][j] = a[i][j];}}res = t;}}cout << res << '\n';for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){ans[i][j] = b[i][j] == 1 ? '.' : 'W';}
// cout << '\n';}}else if(n == 2 && m == 11){ans[0] = "...........";ans[1] = ".W.W.W.W.W.";}else if(n == 2 && m == 12){ans[0] = "............";ans[1] = ".W.W.W.W.W.W";}else if(n == 3 && m == 7){ans[0] = ".......";ans[1] = ".W.W.W.";ans[2] = "..W...W";}else if(n == 3 && m == 8){ans[0] = "........";ans[1] = ".WW.W.W.";ans[2] = "...W...W";}else if(n == 4 && m == 6){ans[0] = ".......";ans[1] = "W.W.W.";ans[2] = ".W.W..";ans[3] = ".....W";}else{ans[0] = ".....";ans[1] = ".W.W.";ans[2] = "..W..";ans[3] = ".W.W.";ans[4] = "....W";}char ans2[30][30];if(!flag){for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){cout << ans[i][j];}cout << '\n';}}else{for(int j = 0; j < m; j++){for(int i = 0; i < n; i++){cout << ans[i][j];}cout << '\n';}}
}
signed main(){ios::sync_with_stdio(0);cin.tie(0);int T = 1;// cin >> T;while(T--){solve();}return 0;
}
