HDU 1010 Tempter of the Bone - (DFS) 奇偶剪枝

/*hdu1010 Tempter of the Bone (DFS)奇偶剪枝: 绕弯的步数-最少的步数（直达） 应该为偶数即可恰好在指定步数到达目的地http://www.cnblogs.com/zhourongqing/archive/2012/04/28/2475684.html --别人的blog*/#include <iostream>#include <cstring>#include <cstdio>#include <queue>#include <algorithm>using namespace std;#define CLR(c,v) (memset(c,v,sizeof(c)))const int M = 10;const char road = '.';const char wall = 'X';const char start = 'S';const char end = 'D';const char inqueue = '$';char *yes = "YES";char *no = "NO";int revise[][2] = {{-1,0},  {0,-1} , {0,1},   {1,0}};int n,m,t;char map[M][M];char *res; // 返回最终结果struct _P{int x,y;_P(int x=-1,int y=-1):x(x),y(y){}bool operator == (const _P& a) const{return ((x==a.x) && (y==a.y));}}S,E; // 起点template <typename _T>inline _T _abs(_T a){return (a>0)?(a):(-a);}void DFS(int step , _P& now){if ( now == E){ if (step == t)res = yes; // 恰好t步能到终点return;}if (step >= t || (_abs(t-step - _abs(now.x-E.x) - _abs(now.y-E.y)) & 1) )return;step++; for (int i = 0 ; i < 4 ; i++){int next_x = now.x + revise[i][0];int next_y = now.y + revise[i][1];if (next_x >= 0 && next_x < n && next_y >= 0 && next_y < m ){ // 没有出界if (map[next_x][next_y] == wall){ // 下一步为墙continue;}char flag  = map[next_x][next_y];map[next_x][next_y] = wall;DFS(step,_P(next_x,next_y));if (res == yes)return;map[next_x][next_y] = flag;}}}int main(){//freopen("in.txt","r",stdin);while(cin >> n >> m >> t && n && m && t){for (int i = 0 ; i < n ; i++){scanf("%s",map[i]);for (int j = 0 ; j < m ; j++){if ( start == map[i][j] )S = _P(i,j);else if(end == map[i][j])E = _P(i,j);}}res = no;map[S.x][S.y] = wall;if (t <= m*n)DFS(0,S);map[S.x][S.y] = start;cout << res << endl;}return 0;}/*一些容易忽略的极限数据,注意剪枝7 7 48D...............................................S7 7 48......D...................................S......7 7 48S...............................................D7 7 48......S...................................D......4 7 27S..........................D3 4 5S.X...X....D4 4 5S.X...X...XD....0 0 0*/