#include <cstdio>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <climits>#include <cstring>#include <cmath>#include <map>#include <set>using namespace std;int n,m,t;int vx[][2] = {{0,1},{1,0},{-1,0},{0,-1}};struct node{int x,y;};char a[10][10];int ans;int flag[10][10];void dfs(node s,int step){node v;if(ans) return;if(a[s.x][s.y] == 'D' && step == t){ans = 1;return ;}if(step >= t)return;for(int i = 0;i < 4;i++){v.x = s.x + vx[i][0];v.y = s.y + vx[i][1];if(a[v.x][v.y] != 'X' && !flag[v.x][v.y]){flag[v.x][v.y] = 1;dfs(v,step+1);flag[v.x][v.y] = 0;if(ans) return;}}} int main(){for(int i = 0;i <= 7+1;i++){for(int j = 0;j <= 7+1;j++){a[i][j] = 'X';}}//freopen("未命名4.cpp","r",stdin);while(scanf("%d%d%d",&n,&m,&t),n||m||t){getchar();node s,e;for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){a[i][j] = getchar();if(a[i][j] == 'S'){s.x = i;s.y = j;}if(a[i][j] == 'D'){e.x = i;e.y = j;}}getchar();}if(abs(s.x-e.x) + abs(s.y – e.y) > t || (s.x + s.y + e.x + e.y + t) % 2 == 1){ //判断狗到终点的曼哈顿距离是否大于t,和狗走到终点的位置是用的步数奇偶性是否相同" << endl;;continue;}ans = 0;memset(flag,0,sizeof(flag));flag[s.x][s.y] = 1;dfs(s,0);if(ans){cout << "YES" << endl;}else{cout << "NO" << endl;}} return 0;}
，有的旅行时为了寻找逝去的年华，重温青春的惆怅。
