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  1. #include<iostream>
  2. #include<stdio.h>
  3. #include<string.h>
  4. using namespace std;
  5. int n, p, q;
  6. int used[26][26];
  7. int pathX[26];
  8. int pathY[26];
  9. int dir[8][2] = { { -2, -1 }, { -2, 1 }, { -1, -2 }, { -1, 2 }, { 1, -2 }, { 1,
  10. 2 }, { 2, -1 }, { 2, 1 } };
  11. int ok;
  12. void dfs(int x, int y, int step) {
  13.     pathX[step - 1] = x;
  14.     pathY[step - 1] = y;
  15.     used[x][y] = 1;
  16.     if (step == p * q) {
  17.         ok = 1;
  18.         return;
  19.     }
  20.     int i;
  21.     for (i = 0; i < 8; i++) {
  22.         int _x = x + dir[i][0];
  23.         int _y = y + dir[i][1];
  24.         if (_x >= 0 && _y >= 0 && _x < p && _y < q && !used[_x][_y]) {
  25.             dfs(_x, _y, step + 1);
  26.         }
  27.     }
  28.     used[x][y] = 0;
  29. }
  30. int main() {
  31.     int sum = 0;
  32.     cin >> n;
  33.     while (n--) {
  34.         cin >> p >> q;
  35.         ok = 0;
  36.         memset(used, 0, sizeof(used));
  37.         memset(pathX, 0, sizeof(pathX));
  38.         memset(pathY, 0, sizeof(pathY));
  39.         printf("Scenario #%d:\n", ++sum);
  40.         dfs(0, 0, 1);
  41.         if (ok != 1) {
  42.             printf("impossible\n");
  43.         } else {
  44.             for (int i = 0; i < p * q; i++) {
  45.                 printf("%c%d", pathY[i] + 'A', pathX[i] + 1);
  46.             }
  47.             printf("\n");
  48.         }
  49.         printf("\n");
  50.     }
  51.     return 0;
  52. }
Success #stdin #stdout 0.01s 2728KB
comments (?)
stdin 
3 
1 1 
2 3 
4 3
stdout 
Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
https://ideone.com/Lfbrh
language:
C++ (gcc 8.3)
created:
12 years ago
visibility:
public

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