hdu 1010 dfs+剪枝

最新推荐文章于 2025-12-20 10:32:27 发布

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原文链接：http://www.cnblogs.com/Xycdada/p/6713586.html

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#数据结构与算法 #java

本文详细解析了一种迷宫逃脱算法，旨在帮助一只狗在有限时间内找到出口。通过深度优先搜索和剪枝策略，算法有效地减少了搜索空间，确保在特定时间点到达门的位置。文章还介绍了两种剪枝方法，包括剩余步数比较和奇偶性检查，以提高算法效率。

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 119601 Accepted Submission(s): 32313

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

剪枝：

第一个：当剩下的步数大于剩下的时间，这时可以直接判断无法到达。

第二个：奇偶性剪枝　　　

　　　　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

　　　0->1 || 1->0 需要奇数步

　　　0->0 || 1->1 需要偶数步

　　　所以我们判断当前位置和D的位置的奇偶，再判断剩余步数的奇偶。

 1 #include<iostream>
 2 
 3 using namespace std;
 4 
 5 int n,m,t,di,dj;
 6 char num[9][9];
 7 bool escape;
 8 int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}};
 9 
10 void dfs(int si,int sj,int cnt)
11 {
12     int i,temp;
13     if(si>n || sj>m || si<=0 || sj<=0)
14         return ;
15     if(cnt==t && si==di && sj==dj)
16         escape=1;
17     if(escape)
18         return ;
19     //奇偶剪枝
20     temp = (t-cnt)-abs(si-di)-abs(sj-dj);
21     if(temp<0 || temp&1)
22         return ;
23     
24     for(i=0;i<4;i++)
25     {
26         if(num[si+dir[i][0]][sj+dir[i][1]]!='X')
27         {
28             num[si+dir[i][0]][sj+dir[i][1]]='X';
29             dfs(si+dir[i][0],sj+dir[i][1],cnt+1);
30             num[si+dir[i][0]][sj+dir[i][1]]='.';
31         }
32     }
33     return ;
34 }
35 int main()
36 {
37     int i,j,si,sj;
38     while(cin>>n>>m>>t,n||m||t)
39     {
40         int wall = 0;
41         for(i=1;i<=n;i++)
42             for(j=1;j<=m;j++)
43             {
44                 cin>>num[i][j];
45                 if(num[i][j]=='S')
46                     si=i,sj=j;
47                 else
48                     if(num[i][j]=='D')
49                         di=i,dj=j;
50                 else
51                     if(num[i][j]=='X')
52                         wall++;
53             }
54         if(n*m-wall<=t)
55         {
56             cout<<"NO"<<endl;
57             continue;
58         }
59         escape=0;
60         num[si][sj]='X';
61         dfs(si,sj,0);
62         if(escape)
63             cout<<"YES"<<endl;
64         else
65             cout<<"NO"<<endl;
66     }
67     return 0;
68 }