HDU 1010 && ZOJ 2110--Tempter of the Bone【DFS && 奇偶剪枝】

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 89408    Accepted Submission(s): 24316

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

 

http://blog.csdn.net/code_pang/article/details/8839432

http://blog.csdn.net/chyshnu/article/details/6171758/

上面是奇偶剪枝的讲解。

题目：有一只狗要吃骨头，结果进入了一个迷宫陷阱，迷宫里每走过一个地板费时一秒，该地板 就会在下一秒塌陷，所以你不能在该地板上逗留。迷宫里面有一个门，只能在特定的某一秒才能打开，让狗逃出去。现在题目告诉你迷宫的大小和门打开的时间，问你狗可不可以逃出去，可以就输出YES,否则NO。

开始没有标记起点，提交的时候wa了一次

ac代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;char map[10][10];int sx, sy, ex, ey;int n, m, t, flag;int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};bool check(int x, int y){    if(x >= 0 && x < n && y >= 0 && y < m && map[x][y] != 'X')        return true;    else        return false;}void dfs(int x, int y, int step){    if(flag)        return ;    if(x ==ex && y == ey && step == t){        flag = 1;        return ;    }    if(step > t)        return ;    int temp = t - step - (abs(x - ex) + abs(y - ey));    if(temp < 0 || temp & 1)        return ;    for(int i = 0; i < 4; ++i){        int fx = x + dir[i][0];        int fy = y + dir[i][1];        if(check(fx, fy)){            map[fx][fy] = 'X';            dfs(fx, fy, step + 1);            map[fx][fy] = '.';        }    }}int main (){    while(scanf("%d%d%d", &n, &m, &t) , n || m || t){        int ans = 0;        for(int i = 0 ; i < n; ++i){            scanf("%s", map[i]);            for(int j = 0 ; j < m; ++j){                if(map[i][j] == 'S'){                    sx = i;                    sy = j;                }                if(map[i][j] == 'D'){                    ex = i;                    ey = j;                }                if(map[i][j] == 'X'){                    ans++;                }            }        }        flag = 0;        map[sx][sy] = 'X';        if(n * m - ans >= t)            dfs(sx, sy, 0);        if(flag)            printf("YES\n");        else            printf("NO\n");    }    return 0;}