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
题目意思：n行m列规模的迷宫，小狗要在t秒内从S点到达D点，问能否完成。
解题思路：搜索的一道很经典的例题。注意两点：
1.需要剪枝，这里有两处剪枝，一个是路径剪枝，一个是奇偶剪枝。路径剪枝一看就明白，关于奇偶剪枝这个给出说明Gate
2.回退过程中现场的恢复。如果当前搜索方向行不通，该搜索过程要结束了，但并不代表其他的搜索方向也行不通，所以回退的时候必须还原到原来的状态，保证其他搜索过程不受影响。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
char maps[9][9];
int n,m,t;
int di,dj;//出口
int flag;//是否成功脱逃
int dir[4][2]= {{0,-1},{0,1},{1,0},{-1,0}}; //四方搜索
void DFS(int x,int y,int cnt)
{
int i,temp;
int a,b;
if (x>n || y>m || x<=0 || y<=0)//越出边界便不搜索
{
return;
}
if (x==di && y==dj && cnt==t) //时间正好的时候才能逃生
{
flag=1;
return;
}
temp=abs(t-cnt)-(abs(di-x)+abs(dj-y));
//计算当前到终点的最短路与还需要的时间差,若小于0则路径剪枝
if (temp<0 || temp%2)//temp如果是奇数的话也要剪枝
{
return;//不符合条件
}
for (i=0; i<4; i++)
{
a=x+dir[i][0];
b=y+dir[i][1];
if (maps[a][b]!='X')
{
maps[a][b]='X';//前进方向！
DFS(a,b,cnt+1);//搜索该点
if (flag)
{
return;
}
maps[a][b]='.';//后退方向！恢复现场！
}
}
return ;
}
int main()
{
int i,j;
int wall;
int si,sj;//起点
while(scanf("%d%d%d",&n,&m,&t)!=EOF)
{
if(n==0&&m==0&&t==0)
{
break;
}
getchar();
wall=0;
for(i=1; i<=n; i++)
{
for(j=1; j<=m; j++)
{
scanf("%c",&maps[i][j]);
if(maps[i][j]=='S')
{
si=i;
sj=j;
}
else if(maps[i][j]=='D')
{
di=i;
dj=j;
}
else if(maps[i][j]=='X')
{
wall++;
}
}
getchar();
}
if(n*m-wall<=t)//路径剪枝，当走完不含墙的迷宫都还不到t时间将不能逃生
{
printf("NO
");
continue;
}
flag=0;
maps[si][sj]='X';//刷为x
DFS(si,sj,0);
if(flag)
{
printf("YES
");
}
else
{
printf("NO
");
}
}
return 0;
}