奇偶性剪枝
若有一地图，将地图的每一个位置有0或1表示（x+y为偶数时 为0 否则为1）：
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
从图中可以看出:任意一个位置周围相邻的必然是与本身值相反的值，也就是说，要想走到与本身值相同的点必然要走偶数步；同理，要想走到与本身相异的值的点必然要走奇数步；
所以，当两个位置的奇偶性相同时（同为0或同为1）若时间为奇数 则必然无法到达；当两个位置的奇偶性不同时（一个为1，另一个为0）若时间为偶数也必不能到达。
判断方法之一：
if (startx+starty+endx+endy+t)%2==1)
printf("无法经过t步从start到达end")
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.
注意一下用cin比较方便，如果用scanf("%c", &ch)的话还要考虑每一行的回车
代码：
#include <iostream>
#include <cmath>
using namespace std;
char maze[10][10];
int startx,starty,endx,endy,t,n,m;
int flag;
int nextt[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
void dfs(int x,int y,int step)
{
if(flag==1) return;
if(maze[x][y]=='X') return;
if(x==endx&&y==endy&&step==t){
flag=1;
return;
}
if(x<1||x>n||y<1||y>m) return;
if(step>t)return;
for(int k=0;k<4;k++)
{
maze[x][y]='X';
dfs(x+nextt[k][0],y+nextt[k][1],step+1);
maze[x][y]='.';
if(flag==1) return;
}
}
int main()
{
int block;
while(cin>>n>>m>>t)
{
block=0;
flag=0;
if(n==0&&m==0&&t==0)break;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>maze[i][j];
if(maze[i][j]=='S'){
startx=i;
starty=j;
continue;
}
if(maze[i][j]=='X'){
block++;
continue;
}
if(maze[i][j]=='D'){
endx=i;
endy=j;
}
}
}
if(t>=(n*m-block)){//剩x个格，只能走x-1步
cout<<"NO"<<endl;
continue;
}
if((endx+startx+starty+endy+t)%2==1){
cout<<"NO"<<endl;
continue;
}
dfs(startx,starty,0);
if(flag==1)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}