hdu 1010 Tempter of the Bone

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 123589    Accepted Submission(s): 33390

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 5S.X...X...XD....3 4 5S.X...X....D0 0 0

 

Sample Output

NOYES

 

•典型的迷宫搜索，熟练掌握该题将具有里程碑式的意义！

•每个block只能走一次

•要求恰好某个给定的时间到达出口

需要用到DFS，但是直接写一个裸的DFS然后XJBB肯定是会TLE，所以我们需要进行一些优化，也就是剪枝

•可以把map看成这样：

•0 1 0 1 0 1

•1 0 1 0 1 0

•0 1 0 1 0 1

•1 0 1 0 1 0

•0 1 0 1 0 1

•从为 0的格子走一步，必然走向为1 的格子

•从为 1的格子走一步，必然走向为0 的格子

•即：

•0->1或1->0必然是奇数步

 0->0走1->1 必然是偶数步

结论：

所以当遇到从 0 走向0 但是要求时间是奇数的，或者，从1走向0但是要求时间是偶数的都可以直接判断不可达！

也就是奇偶性剪枝

#include<stdio.h>#include<math.h>#include<algorithm>int time, atx, aty, n, m;char map[26][26];bool flag;void dfs(int x, int y, int t){    if (x<0 || x >= n || y<0 || y >= m)        return;    if (flag == true || (t == 0 && x == atx&&aty == y))    {        flag = true;        return;    }    int temp = t - abs(x - atx) - abs(y - aty); //剪枝     if (temp<0 || temp & 1)//奇偶性剪枝   与运算 判断奇偶，偶时才行        return;    map[x][y] = 'X';    if (map[x + 1][y] != 'X')    {        dfs(x + 1, y, t - 1);        if (flag)            return;    }    if (map[x][y + 1] != 'X')    {        dfs(x, y + 1, t - 1);        if (flag)            return;    }    if (map[x - 1][y] != 'X')    {        dfs(x - 1, y, t - 1);        if (flag)            return;    }    if (map[x][y - 1] != 'X')    {        dfs(x, y - 1, t - 1);        if (flag)            return;    }    map[x][y] = '.';}int main(){    int i, j, x, y, wall;    while (scanf("%d%d%d", &n, &m, &time) && (n != 0 || m != 0 || time != 0))    {        flag = false;        wall = 0;        for (i = 0; i<n; ++i)        {            scanf("%s", map[i]);            for (j = 0; j<m; ++j)            {                if (map[i][j] == 'S')                {                    x = i;                    y = j;                }                if (map[i][j] == 'D')                {                    atx = i;                    aty = j;                }                if (map[i][j] == 'X')                    wall++;            }        }        dfs(x, y, time);        if (n*m - wall <= time)        {            printf("NO\n");            continue;        }        if (flag)            printf("YES\n");        else            printf("NO\n");    }    return 0;}