杭电ACM1010

Problem Description

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

典型的回溯算法。废话不多说，直接上代码。^_^

#include 
#include       //abs

using namespace std;

const int SIZE = 7;

//定义全局变量,方便下面函数调用
int T, startRow, startCol, doorRow, doorCol, N, M;
char maze[SIZE][SIZE];
bool bmaze[SIZE][SIZE];     //记录是否已走,true表示已走,false表示未走
bool isLived;       //小狗存活标记

bool canMove(int row, int col)    //row-行,col-列
{
    //判断下一个位置是否可走
    if (row >= 0 && row < N && col < M && col >= 0 && !bmaze[row][col])
    {
        bmaze[row][col] = true;
        --T;        //时间减1
        return true;
    }
    return false;
}

bool canLived(int curRow, int curCol)   //判断小狗是否能存活
{
    return curRow == doorRow && curCol == doorCol && T == 0;
}

void resetBmazeAndTime(int row, int col)        //条件不满足，回退，复位前一步
{
    bmaze[row][col] = false;
    ++T;
}

void DFS(int startRow, int startCol)      //start--上一个起点
{
    if (isLived)    //小狗存活，全部递归结束
        return ;

    if (canLived(startRow, startCol))
    {
        isLived = true;
        return ;
    }

    if (canMove(startRow + 1, startCol))     //down
    {
        DFS(startRow + 1, startCol);
        resetBmazeAndTime(startRow + 1, startCol);
    }
    if (canMove(startRow - 1, startCol))     //up
    {
        DFS(startRow -1, startCol);
        resetBmazeAndTime(startRow - 1, startCol);
    }
    if (canMove(startRow, startCol -1))     //left
    {
        DFS(startRow, startCol - 1);
        resetBmazeAndTime(startRow, startCol - 1);
    }
    if (canMove(startRow, startCol + 1))   //right
    {
        DFS(startRow, startCol + 1);
        resetBmazeAndTime(startRow, startCol + 1);
    }
}

int main()
{
    while (cin >> N >> M >> T, N+M+T)
    {
        //初始化
        startRow = 0;
        startCol = 0;
        doorRow  = 0;
        doorCol  = 0;
        isLived = false;
        int block = 0;
        //输入地图
        for (int i = 0; i < N; ++i)
        {
            for (int j = 0; j < M; ++j)
            {
                cin >> maze[i][j];

                if (maze[i][j] == 'S')
                {
                    startRow = i;
                    startCol = j;
                }
                if (maze[i][j] == 'D')
                {
                    doorRow = i;
                    doorCol = j;
                }
                if (maze[i][j] == 'X')
                    ++block;
                if (maze[i][j] == '.' || maze[i][j] == 'D')
                    bmaze[i][j] = false;
                else
                    bmaze[i][j] = true;
            }
        }
        //奇偶剪枝，减法可能出现负数，特别要注意！我之前一直WA就是没注意这里！！
        if (abs(doorRow - startRow + doorCol - startCol) % 2 != T % 2)
        {
            cout << "NO" << endl;
            continue;
        }
        //走的距离>T
        if (abs(doorRow - startRow + doorCol - startCol) > T)
        {
            cout << "NO" << endl;
            continue;
        }
        if (N*M - block < T)    //. 小于 T
        {
            cout << "NO" << endl;
            continue;
        }
        DFS(startRow, startCol);
        if (isLived)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
    return 0;
}