思路：
状压dp。
实现：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 6 const int N = 20; 7 8 char a[N][N]; 9 int n, s, t, ans[N + 5], mp[N][N]; 10 int dp[(1 << N) + 5][N]; 11 int path[(1 << N) + 5][N]; 12 bool ok = false; 13 14 int dfs(int S, int now) 15 { 16 if (dp[S][now] != -1) 17 return dp[S][now]; 18 if (S == (1 << n) - 1) 19 { 20 if (mp[now][s]) 21 { 22 t = now; 23 return ok = 1; 24 } 25 return 0; 26 } 27 for (int i = 0; i < n; i++) 28 { 29 if (ok) 30 continue; 31 if (!(S & (1 << i)) && mp[now][i]) 32 { 33 int res = dfs(S | (1 << i), i); 34 path[S][now] = i; 35 if (res) 36 return dp[S][now] = 1; 37 } 38 } 39 return dp[S][now] = (ok ? -1 : 0); 40 } 41 42 void getPath() 43 { 44 int p = 1 << s; 45 ans[0] = s; 46 int i = 1; 47 for (int j = 0; j < n; j++) 48 { 49 ans[i++] = s = path[p][s]; 50 p |= (1 << s); 51 } 52 } 53 54 int main() 55 { 56 cin >> n; 57 for (int i = 0; i < n; i++) 58 { 59 for (int j = 0; j < n; j++) 60 { 61 cin >> a[i][j]; 62 if (a[i][j] == 'W') 63 mp[i][j] = 1; 64 else if (a[i][j] == 'L') 65 mp[j][i] = 1; 66 } 67 } 68 memset(dp, -1, sizeof(dp)); 69 for (int i = 0; i < n; i++) 70 { 71 s = i; 72 dfs(1 << i, i); 73 if (ok) 74 break; 75 } 76 if (!ok) 77 { 78 cout << "No Solution" << endl; 79 } 80 else 81 { 82 getPath(); 83 for (int i = 0; i < n; i++) 84 { 85 cout << ans[i] + 1; 86 if (i != n - 1) 87 cout << " "; 88 } 89 puts(""); 90 } 91 return 0; 92 }