Tempter of the Bone

秒速五厘米 2022-06-01 01:37 161阅读 0赞

# Tempter of the Bone #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 133935    Accepted Submission(s): 36022**

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 5S.X...X...XD....3 4 5S.X...X....D0 0 0

Sample Output

NOYES

Author

ZHANG, Zheng

思路：刚开始直接搜索，最后提交发现题目理解错了，他要的是狗刚好跑到门刚好开，然后又做发现还是不对，时间超限，百度题解发现还要什么奇偶剪枝，这个剪枝还是不会，这是链接：[点击打开链接][Link 1]

代码：

#include<stdio.h>
    #include<math.h>
    #include<algorithm>
    using namespace std;
    char s[9][9];
    int n,m,t,x,dx,y,dy,flag;//x,y为起始点，dx,dy为终点
    int a[4][2]= {
       {0,1},{0,-1},{1,0},{-1,0}};//上下左右，
    void f(int x,int y,int time)
    {
        int i,ans;
        if(x==dx&&y==dy&&time==t)
            flag=1;
        if(flag)
            return;
        if(x<1||y<1||x>n||y>m)
            return;  //搜索到边界 。
        ans=t-time-abs(dx-x)-abs(dy-y);//奇偶剪枝,超厉害的一个东西。
        if(ans<0||ans&1)//ans&1为判断ans奇偶
            return;
        for(i=0; i<4; i++)
        {
            if(s[x+a[i][0]][y+a[i][1]]!='X')
            {
                s[x+a[i][0]][y+a[i][1]]='X';
                f(x+a[i][0],y+a[i][1],time+1);
                s[x+a[i][0]][y+a[i][1]]='.';
            }
        }
    }
    int main()
    {
        int i,j,k;
        while(scanf("%d%d%d",&n,&m,&t)!=EOF)
        {
            getchar();
            if(n==0&&m==0&&t==0)
                break;
            for(i=1,k=0; i<=n; i++)
            {
                for(j=1; j<=m; j++)
                {
                    scanf("%c",&s[i][j]);
                    if(s[i][j]=='S')
                    {
                        x=i;
                        y=j;
                    }
                    if(s[i][j]=='D')
                    {
                        dx=i;
                        dy=j;
                    }
                    if(s[i][j]=='X')
                        k++;
                }
                getchar();
            }
            if(n*m-k<t)//先直接判断，用最长的时间。如果小于肯定失败
                printf("NO\n");
            else
            {
                flag=0;
                s[x][y]='X';//标记已经访问过了。
                f(x,y,0);
                if(flag==1)
                    printf("YES\n");
                else
                    printf("NO\n");
            }
        }
        return 0;
    }

[Link 1]: https://baike.baidu.com/item/奇偶剪枝#2