pid=1010_Tempter of the Bone

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最新推荐文章于 2025-09-12 19:14:19 发布

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分类专栏： HDU 文章标签：算法 bfs dfs

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本文介绍了一道关于迷宫逃脱的编程题目，采用深度优先搜索（DFS）算法解决狗狗在限定时间内找到出口的问题。通过剪枝优化提高算法效率。

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 65317 Accepted Submission(s): 17830

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

自我总结：还太弱，bfs，dfs等算法理解了，但是不能使用，悲哀！！

#include<iostream>
#include<cstring>
using namespace std;

#define N 10
int n,m,t,end_i,end_j;
bool visited[N][N],flag,ans;
char map[N][N];

int abs(int a,int b)
{
    if(a<b) return b-a;
    else return a-b;
}

void DFS(int i,int j,int c)
{
    if(flag) return ;
    if(c>t) return ;    
    if(i<0||i>=n||j<0||j>=m) {return ;}
    if(map[i][j]=='D'&&c==t) {flag=ans=true; return ;}
    int temp=abs(i-end_i)+abs(j-end_j);
    temp=t-temp-c;
    if(temp&1) return ;//奇偶剪枝

    if(!visited[i-1][j]&&map[i-1][j]!='X') 
    {
        visited[i-1][j]=true;
        DFS(i-1,j,c+1);
        visited[i-1][j]=false;
    }
    if(!visited[i+1][j]&&map[i+1][j]!='X') 
    {
        visited[i+1][j]=true;
        DFS(i+1,j,c+1);
        visited[i+1][j]=false;
    }
    if(!visited[i][j-1]&&map[i][j-1]!='X') 
    {
        visited[i][j-1]=true;
        DFS(i,j-1,c+1);
        visited[i][j-1]=false;
    }
    if(!visited[i][j+1]&&map[i][j+1]!='X') 
    {
        visited[i][j+1]=true;
        DFS(i,j+1,c+1);
        visited[i][j+1]=false;
    }
}

int main_1010_tempter_of_bone()
{
    int i,j,x,y,k;
    while(cin>>m>>n>>t&&(m||n||t))
    {
        memset(visited,false,sizeof(visited));
        k=0;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                cin>>map[i][j];
                if(map[i][j]=='S')
                {
                    x=i;y=j;
                    visited[i][j]=true;
                }
                if(map[i][j]=='D')
                {
                    end_i=i;end_j=j;
                }
                if(map[i][j]=='X')k++;
            }
        }
        ans=flag=false;
        if(n*m-k-1>=t) DFS(x,y,0);

        if(ans) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}

转自：http://www.cnblogs.com/zhourongqing/archive/2012/04/28/2475684.html