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要解决这个问题，我们需要找到一个骑士的旅行路径，使其访问棋盘上的每个格子恰好一次。骑士的移动方式是每一步移动两格在一个方向，然后一格在另一个方向，或者相反方向。我们的目标是找到一个字典序最小的路径，如果不存在这样的路径，则返回“impossible”。

方法思路

解决代码

#include 
   
    
#include 
    
     
#include 
     
      
#include 
      
       
using namespace std;
int dx[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dy[] = {-1, 1, -2, 2, -2, 2, -1, 1};
int a, b;
int chessboard_size = a * b;
bool visited[chessboard_size + 1][chessboard_size + 1];
int path[chessboard_size + 1];
int current_step = 0;
bool found = false;
void dfs(int x, int y, int step) {
    path[step] = x;
    visited[x][y] = true;
    if (step == chessboard_size) {
        found = true;
        return;
    }
    for (int i = 0; i < 8; ++i) {
        int nx = x + dx[i];
        int ny = y + dy[i];
        if (nx >= 1 && nx <= b && ny >= 1 && ny <= a && !visited[nx][ny]) {
            if (path[step + 1] == 0 || (path[step + 1] > nx + (ny - 1) / 10 * 10 + 1)) {
                dfs(nx, ny, step + 1);
            }
        }
    }
}
int main() {
    int n;
    scanf("%d", &n);
    for (int cas = 1; cas <= n; ++cas) {
        found = false;
        memset(visited, 0, sizeof(visited));
        memset(path, 0, sizeof(path));
        scanf("%d %d", &a, &b);
        for (int y = 1; y <= a; ++y) {
            for (int x = 1; x <= b; ++x) {
                if (!visited[x][y]) {
                    if (!found) {
                        if (a * b == 1) {
                            path[1] = x;
                            visited[x][y] = true;
                            found = true;
                        } else {
                            for (int i = 0; i < 8; ++i) {
                                int nx = x + dx[i];
                                int ny = y + dy[i];
                                if (nx >= 1 && nx <= b && ny >= 1 && ny <= a) {
                                    if (path[1] == 0 || (path[1] > nx + (ny - 1) / 10 * 10 + 1)) {
                                        path[1] = nx;
                                        visited[nx][ny] = true;
                                        dfs(nx, ny, 1);
                                        if (found) break;
                                        visited[nx][ny] = false;
                                    }
                                }
                            }
                            if (found) break;
                        }
                    }
                }
            }
            if (found) break;
        }
        if (found) {
            for (int i = 1; i <= chessboard_size; ++i) {
                char c = 'A' + (path[i] - 1) / 10;
                int num = path[i] - (path[i] - 1) / 10 * 10 - 1;
                printf("%c%d", c, num);
            }
            printf("\n");
        } else {
            printf("impossible\n");
        }
        printf("\n");
    }
    return 0;
}
      
     
    
   

代码解释

通过这种方法，我们可以高效地解决问题，并找到字典序最小的骑士旅行路径。

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