Tempter of the Bone
The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.
The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.
Input
The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:
'X': a block of wall, which the doggie cannot enter;
'S': the start point of the doggie;
'D': the Door; or
'.': an empty block.
The input is terminated with three 0's. This test case is not to be processed.
Output
For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.
Sample Input
4 4 5
S.X.
..X.
..XD
....
3 4 5
S.X.
..X.
...D
0 0 0
Sample Output
NO
YES
题目类型：DFS +剪枝
算法分析：原本是一道简单的DFS题目，但是需要注意搜索的效率，如果直接使用DFS会超时，所以重点是使用剪枝，剪枝的地方是对于搜索点到目标点的距离所用时间已经大于实际提供的时间，就不再向下搜索。另一个重要的剪枝(只有前一个剪枝仍会TLE)是前两个时间之差必须是偶数，即意味着两个时间必须是同奇偶的才有可能找到结果，事实上第二个剪枝的所剪掉的无用搜索更多，只用第二个剪枝程序仍能AC
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#include <iostream> #include <fstream> #include <cstring> #include <cmath> using namespace std; const int dx[] = {0, 0, -1, 1}; const int dy[] = {-1, 1, 0, 0}; const int maxn = 36 + 6; char ans[maxn][maxn]; bool visited[maxn][maxn]; int n, m, t; int s_x, s_y, e_x, e_y; bool is_valid; bool OkMove (int x, int y) { if (x >= 0 && x <= n - 1 && y >= 0 && y <= m - 1 && ans[x][y] != 'X' && !visited[x][y]) return true; return false; } void DFS (int x, int y, int cnt) { if (x == e_x && y == e_y && cnt == t) { is_valid = true; return ; } int temp = (t - cnt) - abs (x - e_x) - abs (y - e_y); if (temp < 0 || temp % 2) return ; int i; for (i = 0; i < 4; i++) { if (OkMove (x + dx[i], y + dy[i])) { int temp_x = x + dx[i], temp_y = y + dy[i]; visited[temp_x][temp_y] = true; DFS (temp_x, temp_y, cnt + 1); visited[temp_x][temp_y] = false; } } } int main() { // ifstream cin ("aaa.txt"); while (cin >> n >> m >> t) { if (n == 0 && m == 0 && t == 0) break; int i, j; for (i = 0; i < n; i++) for (j = 0; j < m; j++) { cin >> ans[i][j]; if (ans[i][j] == 'S') { s_x = i; s_y = j; } if (ans[i][j] == 'D') { e_x = i; e_y = j; } } is_valid = false; memset (visited, false, sizeof (visited)); visited[s_x][s_y] = true; DFS (s_x, s_y, 0); if (is_valid) cout << "YES" << endl; else cout << "NO" << endl; } return 0; } |
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