Description
The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and must wander through the maze facing zombies, chainsaw-wielding psychopaths, hippies, and other terrors on their quest to find the exit.
One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.)
As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.
One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.)
As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.
Input
Input to this problem will begin with a line containing a single integer n indicating the number of mazes. Each maze will consist of one line with a width, w, and height, h (3 <= w, h <= 40), followed by h lines of w characters each that represent the maze layout. Walls are represented by hash marks ('#'), empty space by periods ('.'), the start by an 'S' and the exit by an 'E'.
Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#').
You may assume that the maze exit is always reachable from the start point.
Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#').
You may assume that the maze exit is always reachable from the start point.
Output
For each maze in the input, output on a single line the number of (not necessarily unique) squares that a person would visit (including the 'S' and 'E') for (in order) the left, right, and shortest paths, separated by a single space each. Movement from one square to another is only allowed in the horizontal or vertical direction; movement along the diagonals is not allowed.
Sample Input
2 8 8 ######## #......# #.####.# #.####.# #.####.# #.####.# #...#..# #S#E#### 9 5 ######### #.#.#.#.# S.......E #.#.#.#.# #########
Sample Output
37 5 5 17 17 9
这题想了好久，要求三个数，一个是靠左先 一个靠右先，一个最短到终点的距离，其中最短距离很好求，简单bfs就行了
就是其他两个不太好想。本想着有什么简单的方法的，可结果发了两三天都没想出来，最后就在纸上画它行走的方向图，
发现每次走完之后所改变的方向都是对应的，即一个方向图向某一方向同时移动一位，构成对应的方向关系- -虽然想到
这里但是还是没什么简单的方法。。DT时。。决定用枚举写了- -结果成了现在的冗长的代码。。
View Code
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <queue> 6 using namespace std; 7 const int N=45; 8 const int east=0; 9 const int north=1; 10 const int west=2; 11 const int south=3; 12 int dx[]={-1,0,1,0}; 13 int dy[]={0,1,0,-1}; 14 struct road 15 { 16 int x,y; 17 int step; 18 }start,end,mid,t; 19 int vis[N][N]; 20 int map[N][N]; 21 int w,h; 22 int bfs() 23 { 24 int i; 25 memset(vis,0,sizeof(vis)); 26 queue<road>q; 27 q.push(start); 28 vis[start.x][start.y]=1; 29 while(!q.empty()) 30 { 31 mid=q.front(); 32 q.pop(); 33 if(mid.x==end.x&&mid.y==end.y) 34 { 35 end.step=mid.step; 36 return end.step; 37 } 38 for(i=0;i<4;i++) 39 { 40 t.x=mid.x+dx[i]; 41 t.y=mid.y+dy[i]; 42 if(t.x>=0&&t.x<h&&t.y>=0&&t.y<w&&map[t.x][t.y]&&!vis[t.x][t.y]) 43 { 44 vis[t.x][t.y]=1; 45 t.step=mid.step+1; 46 if(t.x==end.x&&t.y==end.y) 47 return t.step; 48 q.push(t); 49 } 50 } 51 } 52 } 53 int left(int x,int y,int d) 54 { 55 int cnt=1; 56 while(map[x][y]!=2) 57 { 58 switch(d) 59 { 60 case north: 61 if(y-1>=0&&map[x][y-1]) 62 y--,d=west; 63 else if(x-1>=0&&map[x-1][y]) 64 x--,d=north; 65 else if(y+1<w&&map[x][y+1]) 66 y++,d=east; 67 else if(x+1<h&&map[x+1][y]) 68 x++,d=south; 69 cnt++; 70 break; 71 case west: 72 if(x+1<h&&map[x+1][y]) 73 x++,d=south; 74 else if(y-1>=0&&map[x][y-1]) 75 y--,d=west; 76 else if(x-1>=0&&map[x-1][y]) 77 x--,d=north; 78 else if(y+1<w&&map[x][y+1]) 79 y++,d=east; 80 cnt++; 81 break; 82 case south: 83 if(y+1<w&&map[x][y+1]) 84 y++,d=east; 85 else if(x+1<h&&map[x+1][y]) 86 x++,d=south; 87 else if(y-1>=0&&map[x][y-1]) 88 y--,d=west; 89 else if(x-1>=0&&map[x-1][y]) 90 x--,d=north; 91 cnt++; 92 break; 93 case east: 94 if(x-1>=0&&map[x-1][y]) 95 x--,d=north; 96 else if(y+1<w&&map[x][y+1]) 97 y++,d=east; 98 else if(x+1<h&&map[x+1][y]) 99 x++,d=south; 100 else if(y-1>=0&&map[x][y-1]) 101 y--,d=west; 102 cnt++; 103 break; 104 } 105 } 106 return cnt; 107 } 108 int right(int x,int y,int d) 109 { 110 int cnt=1; 111 while(map[x][y]!=2) 112 { 113 switch(d) 114 { 115 case north: 116 if(y+1<w&&map[x][y+1]) 117 y++,d=east; 118 else if(x-1>=0&&map[x-1][y]) 119 x--,d=north; 120 else if(y-1>=0&&map[x][y-1]) 121 y--,d=west; 122 else if(x+1<h&&map[x+1][y]) 123 x++,d=south; 124 cnt++; 125 break; 126 case west: 127 if(x-1>=0&&map[x-1][y]) 128 x--,d=north; 129 else if(y-1>=0&&map[x][y-1]) 130 y--,d=west; 131 else if(x+1<h&&map[x+1][y]) 132 x++,d=south; 133 else if(y+1<w&&map[x][y+1]) 134 y++,d=east; 135 cnt++; 136 break; 137 case south: 138 if(y-1>=0&&map[x][y-1]) 139 y--,d=west; 140 else if(x+1<h&&map[x+1][y]) 141 x++,d=south; 142 else if(y+1<w&&map[x][y+1]) 143 y++,d=east; 144 else if(x-1>=0&&map[x-1][y]) 145 x--,d=north; 146 cnt++; 147 break; 148 case east: 149 if(x+1<h&&map[x+1][y]) 150 x++,d=south; 151 else if(y+1<w&&map[x][y+1]) 152 y++,d=east; 153 else if(x-1>=0&&map[x-1][y]) 154 x--,d=north; 155 else if(y-1>=0&&map[x][y-1]) 156 y--,d=west; 157 cnt++; 158 break; 159 } 160 } 161 return cnt; 162 } 163 int main() 164 { 165 int t,i,j,dir; 166 char s[50]; 167 scanf("%d",&t); 168 while(t--) 169 { 170 scanf("%d%d",&w,&h); 171 for(i=0;i<h;i++) 172 { 173 scanf("%s",s); 174 for(j=0;s[j];j++) 175 { 176 if(s[j]=='.'||s[j]=='E') 177 { 178 if(s[j]=='E') 179 map[i][j]=2; 180 else 181 map[i][j]=1; 182 } 183 if(s[j]=='#'||s[j]=='S') 184 map[i][j]=0; 185 if(s[j]=='S') 186 { 187 start.x=i; 188 start.y=j; 189 start.step=1; 190 } 191 if(s[j]=='E') 192 { 193 end.x=i; 194 end.y=j; 195 } 196 } 197 } 198 if(start.x==0) dir=south; 199 else if(start.x==h-1) dir=north; 200 else if(start.y==0) dir=east; 201 else if(start.y==w-1) dir=west; 202 printf("%d %d %d\n",left(start.x,start.y,dir),right(start.x,start.y,dir),bfs()); 203 } 204 return 0; 205 }