一开始的思路是,先走驴子,之后走老虎,每次走的时候记录他们的达到该位置的步数,如果一样就可以相遇,特殊处理末尾。
但是没办法过,之后索性让他们一起走,一起修改移动的坐标,结果过了。。。
真心不知道错哪了。。
11662502 | 2014-09-16 00:43:01 | Accepted | 4740 | 31MS | 8300K | 2547 B | G++ | KinderRiven |
#include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<string> #include<vector> #include<stack> #include<set> #include<algorithm> #include<map> #include<sstream> #include<cmath> #include<queue> #define INF (1 << 30) #define eps (1e-10) #define _PI acos(-1.0) using namespace std; /*========================================= =========================================*/ #define MAXD 1000 + 10 int n; int dir[4][2] = {{0,1},{1,0},{0,-1},{-1,0}}; int aa , bb; int vis1[MAXD][MAXD],vis2[MAXD][MAXD]; bool solve(int x1,int y1,int d1,int x2,int y2,int d2){ memset(vis1,0,sizeof(vis1)); memset(vis2,0,sizeof(vis2)); vis1[x1][y1] = 1; vis2[x2][y2] = 1; int _x1 ,_y1; int _x2 ,_y2; while(aa || bb){ if(x1 == x2 && y1 ==y2){ printf("%d %d\n",x1,y1); return true; } if(aa){ _x1 = x1 + dir[d1][0]; _y1 = y1 + dir[d1][1]; if(_x1 >= 0 && _x1 < n && _y1 >= 0 && _y1 < n && !vis1[_x1][_y1]){ vis1[_x1][_y1] = 1; } else{ d1 = (d1 + 1) % 4; _x1 = x1 + dir[d1][0]; _y1 = y1 + dir[d1][1]; if(_x1 >= 0 && _x1 < n && _y1 >= 0 && _y1 < n && !vis1[_x1][_y1]){ vis1[_x1][_y1] = 1; } else{ aa = 0; } } } if(bb){ _x2 = x2 + dir[d2][0]; _y2 = y2 + dir[d2][1]; if(_x2 >= 0 && _x2 < n && _y2 >= 0 && _y2 < n && !vis2[_x2][_y2]){ vis2[_x2][_y2] = 1; } else{ d2 = (d2 + 3) % 4; _x2 = x2 + dir[d2][0]; _y2 = y2 + dir[d2][1]; if(_x2 >= 0 && _x2 < n && _y2 >= 0 && _y2 < n && !vis2[_x2][_y2]){ vis2[_x2][_y2] = 1; } else{ bb = 0; } } } if(aa) {x1 = _x1; y1 = _y1;} if(bb) {x2 = _x2; y2 = _y2;} } return false; } int main(){ while(scanf("%d",&n) && n){ int x1,y1,x2,y2,d1,d2; aa = 1; bb = 1; scanf("%d%d%d",&x1,&y1,&d1); scanf("%d%d%d",&x2,&y2,&d2); if(!solve(x1,y1,d1,x2,y2,d2)) printf("-1\n"); } return 0; }