[POJ] 2488.A Knight's Journey

题目传送门
题意：日字走，一次走完给定的p*q棋盘，记录步骤，输出
思路：dfs + 回溯

#include <algorithm>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int p, q;
int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int chess[26][26];
bool flag;
struct step {
  int x;
  char y;
} path[26];

void dfs(int x, int y, int counts) {
  path[counts].x = x;
  path[counts].y = y + 'A' - 1;
  int x1, y1;
  if (counts == p * q) {
    flag = true;
    for (int i = 1; i <= p * q; i++) printf("%c%d", path[i].y, path[i].x);
    printf("
");
  }
  if (flag) return;

  for (int i = 0; i < 8; i++) {
    x1 = x + dx[i];
    y1 = y + dy[i];

    if (x1 > 0 && x1 <= p && y1 > 0 && y1 <= q && !chess[x1][y1]) {
      chess[x1][y1] = 1;
      dfs(x1, y1, counts + 1);
      chess[x1][y1] = 0;
    }
  }
}

int main() {
  int n;
  scanf("%d", &n);
  int c = 1;
  while (n--) {
    memset(chess, 0, sizeof(chess));
    flag = false;
    scanf("%d%d", &p, &q);
    printf("Scenario #%d:
", c++);
    chess[1][1] = 1;
    dfs(1, 1, 1);
    if (!flag) printf("impossible
");
    printf("
");
  }
  system("pause");
  return 0;
}