Hdu1010 Tempter of the Bone（DFS+剪枝） 2016-05-06 09:12 432人阅读 评论(0) 收藏

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 102071    Accepted Submission(s): 27649

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

———————————————————————————————

题目的意思是从地图的S点搜到D点，‘.’为路‘X’为墙，问能否正好在t秒时到达终点。

处理方法是直接DFS搜索，加上剪枝。

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

int m, n, t;
char mp[10][10];
int dir[4][2] = { { 1, 0 }, { -1, 0 }, { 0, -1 }, { 0, 1 } };
bool escape;

bool check(int x, int y)
{
    if (x < 0 || x >= m||y < 0 || y >= n||mp[x][y] == 'X')
        return 0;
    return 1;
}

void dfs(int si,int sj,int di,int dj,int cnt)
{
    if (si == di&&sj == dj&&cnt == t)
    {
        escape = 1;
        return;
    }
    int tmp = t-cnt-abs(di - si) - abs(dj - sj);

    if (tmp < 0 || tmp % 2)
        return;

    for (int i = 0; i < 4; i++)
    {
        int xx = si + dir[i][0];
        int yy = sj + dir[i][1];
        if (check(xx, yy))
        {
            mp[xx][yy] = 'X';
            dfs(xx, yy, di, dj, cnt + 1);
            mp[xx][yy] = '.';
            if (escape == 1)
                return;
        }
    }
}

int main()
{
    int si, sj, di, dj;
    while (~scanf("%d%d%d", &m, &n, &t))
    {
        if (m == 0 || n == 0 || t == 0)
            break;
        int wall = 0;
        for (int i = 0; i < m; i++)
        {
            scanf("%s", &mp[i]);
            for (int j = 0; j < n; j++)
            {
                if (mp[i][j] == 'S')
                {
                    si = i;
                    sj = j;
                }
                if (mp[i][j] == 'D')
                {
                    di = i;
                    dj = j;
                }
                if (mp[i][j] == 'X')
                    wall++;
            }

        }
        if (t > m*n - wall)
        {
            printf("NO\n");
            continue;
        }
        escape = 0;
        mp[si][sj] = 'X';
        dfs(si, sj, di, dj, 0);
        if (escape == 1)
            printf("YES\n");
        else
            printf("NO\n");

    }
    return 0;
}