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): 57941    Accepted Submission(s): 15654

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

    题意：

    给出N,M,T（N和M范围1到7，T为50）。代表有一个地图大小有N行M列，S为起点，D为终点，问是否能在T步从S走到D。（必须为T步，不能少于也不能大于）。

 

    思路：

    DFS。但是单纯的DFS会超时，所以要进行剪枝。奇偶剪枝。

 

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
int endx,endy,t,temp,n,m,ans;
char map[10][10];
int vis[10][10],dir[4][2]={-1,0,0,-1,1,0,0,1};

void dfs(int x,int y)
{
    vis[x][y]=1;
 //   printf("x=%d y=%d\n",x,y);
    if(ans>t||temp) return;
//如果统计步数已经大于所要求步数，或者已经找到就不继续搜索了
    if(ans==t&&x==endx&&y==endy)
    {
        temp=1;
        return;
    }
//    if((abs(endx-x)+abs(endy-y))%2!=((t-ans)%2)) return;
    for(int i=0;i<4;i++)
    {
        int nx,ny;
        nx=x+dir[i][0];
        ny=y+dir[i][1];
        if(nx>=1&&ny>=1&&nx<=n&&ny<=m&&!vis[nx][ny]&&map[nx][ny]!='X')
        {
            ans++;
            dfs(nx,ny);
            vis[nx][ny]=0;
            ans--;
        }
    }
}

int main()
{
 //   freopen("text.in","r",stdin);
    int w;
    while(scanf("%d%d%d",&n,&m,&t)!=EOF&&(n+m+t))
    {
        int stax,stay;
        temp=0,ans=0,w=0;
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            {
            scanf(" %c",&map[i][j]);
            if(map[i][j]=='S') stax=i,stay=j;
            if(map[i][j]=='D') endx=i,endy=j;
            if(map[i][j]=='X') w++;
            }
//一开始判断一次就够了
        if((abs(endx-stax)+abs(endy-stay))%2!=(t%2))
        {
            printf("NO\n");
            continue;
        }
//奇偶剪枝
        if(n*m<w+t)
        {
            printf("NO\n");
            continue;
        }
        dfs(stax,stay);
        if(temp) printf("YES\n");
        else     printf("NO\n");
    }
    return 0;
}