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.
'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
方法一：
#include<stdio.h>
#include<string.h>
#include<math.h>
char map[10][10];
int n,m,t,di,dj,wall,flag;
int next[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
void dfs(int xi,int xj,int count)
{
int i,tem;
if( xi > n || xj > m|| xi <=0 || xj <=0)
{
return;
}
if(count==t && xi==di && xj==dj)
{
flag=1;
}
if(flag == 1)
return;
tem = ( t - count) - abs ( xi - di) - abs( xj - dj);
if(tem < 0 || tem%2!=0 )
return ;
for(i=0;i<4;i++)
{
if(map[xi+next[i][0]][xj+next[i][1]]!='X')
{
map[xi+next[i][0]][xj+next[i][1]]='X';
dfs(xi+next[i][0],xj+next[i][1],count+1);
map[xi+next[i][0]][xj+next[i][1]]='.';
}
}
return ;
}
int main()
{
int i,j,si,sj;
while(scanf("%d%d%d%*c",&n,&m,&t),n!=0&&m!=0&&t!=0)
{
wall=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
scanf("%c",&map[i][j]);
if(map[i][j]=='S')
si=i,sj=j;
if(map[i][j]=='D')
di=i,dj=j;
if(map[i][j]=='X')
wall++;
}
getchar();
}
if(n*m-wall<=t)
{
printf("NO\n");
continue;
}
flag=0;
map[si][sj]='X';
dfs(si,sj,0);
if(flag==1)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
方法二：
#include<stdio.h>
#include<math.h>
#include<string.h>
char map[10][10];
int xi,xj,di,dj,n,m,t;
int next[4][2]={{1,0},{-1,0},{0,1},{0,-1}},book[10][10],flag;
void dfs(int xi,int xj,int step)
{
int i,ti,tj;
if(xi==di&&xj==dj)
{
if(t==step)
flag=1;
return;
}
if(step>=t)
return;
if(map[xi][xj]!='X')
{
for(i=0;i<4;i++)
{
ti=xi+next[i][0];
tj=xj+next[i][1];
if(map[ti][tj]!='X'&&ti>=0&&ti<n&&tj>=0&&tj<m&&book[ti][tj]!=1)
{
book[ti][tj]=1;
dfs(ti,tj,step+1);
book[ti][tj]=0;
if(flag)
return;
}
}
}
}
int main()
{
while(scanf("%d%d%d",&n,&m,&t),n!=0&&m!=0&&t!=0)
{
int i,step=0,j;
for(i=0;i<n;i++)
{
getchar();
for(j=0;j<m;j++)
{
scanf("%c",&map[i][j]);
if(map[i][j]=='S')
{
xi=i;
xj=j;
}
if(map[i][j]=='D')
{
di=i;
dj=j;
}
}
}
getchar();
memset(book,0,sizeof(book));
if(abs(xi-di)+abs(xj-dj)>t||(xi+xj+di+dj+t)%2==1)
{
printf("NO\n");
continue;
}
book[xi][xj]=1;
flag=0;
dfs(xi,xj,step);
if(flag==1)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
以上两种方法都用到了奇偶剪枝！！！！！
那什么是奇偶剪枝呢！
其实它的理解很绕，内容有介绍起来很多很烦，但是大家要仔细看，并且去理解它。
先看这样的一个矩阵：
矩阵形式如下：
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
从为 0 的格子走一步，必然走向为 1 的格子 。
从为 1 的格子走一步，必然走向为 0 的格子 。
由上面的可以得到以下的结论：
1.从 0 走向 1 （或 1 走向 0 ）必然是奇数步。
2.从 0 走向 0（ 或 1 走向 1 ）必然是偶数步。
So
当遇到下面两种情况时，都可以直接判断不可达！
- 从 0 走向 0 ，要求时间（所剩的时间）是奇数的。
- 从 1 走向 0 ，要求时间（所剩的时间）是偶数的。
本题的两个解法：
解法1：
tem=(t-count)-abs(xi-di)-abs(xj-dj);
if(tem < 0 || tem%2!=0 )
return;
这是它的核心代码。
令 最少需要的步数为a=abs(xi-di)+abs(xi-xj) ，剩下的步数为b=t-count 。（这里就这样表示了，都懂啊！@-@）
则
如果a为奇数，则一定是 从 0 走向 1 （或 1 走向 0）。
如果a为偶数，则一定是 从 0 走向 0 （或 1 走向 1）。
如果b为奇数，则一定是 从 0 走向 1 （或 1 走向 0）。
如果b为偶数，则一定是 从 0 走向 0 （或 1 走向 1）。
因为
偶数-奇数 = 奇数，奇数-偶数 = 奇数，奇数-奇数= 偶数，偶数-偶数=偶数。
所以
b-a得到结果直接判断奇偶就行了。
解法2的代码
里面用一个book数组来标记这个地方是否走过。其他的问题和解法1 大同小异。耐下心来好好分析就知道啦！！！！！
解释了这么多不知道你们是否能明白。。。。。。。。。。。。。。
有问题留言吧！！！！