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): 23728    Accepted Submission(s): 6527

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

 

 

Author

ZHANG, Zheng

 

 

Source

ZJCPC2004

 

 

Recommend

 

 

尼条题目我WA = 47次， TLE = 3次。

其实尼条题目既精粹就在于奇偶剪枝，同埋判断剩余可行格数。

 

剩余格数可以理解，但系咩系奇偶剪枝？baidu好多解答，呢度我吾讲啦{= =}。

 

呢个博文重点讲一D小技巧，同埋要注意既地方。

下面先贴代码：

 

4266151

2011-07-27 10:20:41

Accepted

1010

46MS

288K

1132 B

C++

10SGetEternal{（。）（。）}!

 

#include <iostream>   
using namespace std;  
#define MAXI 11   
#define ABS(x) ((x) > 0? (x): -(x))   
  
char maz[MAXI][MAXI];  
int  n, m, t, sx, sy, ex, ey, blo;  
int  mov[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};  
  
bool DFS(int x, int y, int st)  
{  
    int dt = t - st - ABS(ex - x) - ABS(ey - y);  
    int i, tx, ty;  
    char tc = maz[x][y];  
  
    if (st == t && x == ex && y == ey) return 1;  
    if (dt < 0 || dt % 2 == 1) return 0;  
  
    maz[x][y] = 'X';  
    for (i = 0; i < 4; i++)  
    {  
        tx = x + mov[i][0];  
        ty = y + mov[i][1];  
        if (0 <= tx && tx < n && 0 <= ty && ty < m && maz[tx][ty] != 'X')  
            if (DFS(tx, ty, st + 1)) return 1;  
    }  
    maz[x][y] = tc;  
  
    return 0;  
}  
  
int main()  
{  
    int i, j;  
  
    while (cin >> n >> m >> t, n || m || t)  
    {  
        for (blo = i = 0; i < n; i++)  
        {  
            //getchar();   
            for (j = 0; j < m; j++)  
            {  
                cin >> maz[i][j];  
                //scanf("%c", maz[i] + j);   
                if (maz[i][j] == '.') blo++;  
                if (maz[i][j] == 'S')  
                { sx = i; sy = j; }  
                if (maz[i][j] == 'D')  
                { ex = i; ey = j; blo++; }  
            }  
        }  
        if (blo >= t && DFS(sx, sy, 0)) printf("YES\n");  
        else printf("NO\n");  
    }  
  
    return 0;  
}  

 
首先重点讲一下WA杀手getchar()

我之前一直无留意呢个地方…………用getchar配合scanf其实要系测试数据无多余空行前提下可以用，杭电测试数据……

所以我就WA好多次都唔知死系边度……改cin之后就AC鸟。

TLE有两次系因为我用maz[x][y] == 'D'判断搜索结束，其实甘样系有问题，因为我地通常改标记会原地改，搜索完一轮之后就原地复原，

不过因为尼条题目特性系要t == tim同时到达D先符合条件，所以有可能某次搜索到达左D但系t > tim，而且呢轮搜索完结之后，又唔记得

复原'D'而系直接赋'.'，所以之后就搵唔到D啦。解决呢个问题有两个方法：一、用坐标判断。二、用一个变量储存标记前坐标。

 

就系甘样浪费左一日