hdu 1010 Tempter of the Bone

最新推荐文章于 2023-10-20 20:27:55 发布

XiaoYaoII

最新推荐文章于 2023-10-20 20:27:55 发布

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分类专栏： ACM

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本文介绍了一款迷宫逃脱游戏的问题背景与解决方案。通过深度优先搜索（DFS）算法及剪枝策略，解决了狗仔在限定时间内从迷宫中逃出的问题。文中详细解释了输入输出格式，并给出了一段实现该算法的C++代码。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1010

题目描述：

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

题目分析：

dfs+剪枝，要注意的地方详见注释

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define LIM (nx>=1&&nx<=n&&ny>=1&&ny<=m&&!v[nx][ny])
const int maxn=8;
int v[maxn][maxn],much[maxn][maxn];
char s[maxn][maxn];
int n,m,t,sign;
int detx,dety;
int stax,stay;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
void dfs(int x,int y,int step)
{
    if(x==detx&&y==dety)
    {
        if(step==t)
            sign=1;//找到符合条件
        return;
    }
    if(step>=t)
        return;
    for(int i=0;i<4;i++)
    {
        int nx=x+dx[i],ny=y+dy[i];
        if(LIM)
        {
            v[nx][ny]=1;
            dfs(nx,ny,step+1);
            v[nx][ny]=0;
            if(sign)//只要找到一个满足条件的，以后的就不用再看了
                return;
        }
    }
}
int main()
{
    while(1)
    {
        scanf("%d%d%d",&n,&m,&t);
        if(!(m||n||t))
            break;
        sign=0;
        memset(v,0,sizeof(v));
        for(int i=1;i<=n;i++)
        {
                scanf("%s",&s[i][1]);
                for(int j=1;j<=m;j++)
                {
                    if(s[i][j]=='X')
                        v[i][j]=1;
                    else if(s[i][j]=='S')
                        {
                            stax=i;
                            stay=j;
                            v[i][j]=1;
                        }
                    else if(s[i][j]=='D')
                    {
                        detx=i;
                        dety=j;
                    }
                }
        }
        if((abs(detx-stax)+abs(dety-stay)>t)||((detx+dety+stax+stay+t)%2==1))//这一步剪枝特别重要，可以减少700~800ms，(detx+dety+stax+stay+t)%2==1这一步看不懂的小伙伴可以去aoj667 看看，会找到灵感
        {
            printf("NO\n");
            continue;
        }
        dfs(stax,stay,0);
        if(sign)
            printf("YES\n");
        else
            printf("NO\n");
    }
}