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poj 3083 dfs，bfs
Children of the Candy Corn
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uDescription
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
28 8#########......##.####.##.####.##.####.##.####.##...#..##S#E####9 5##########.#.#.#.#S.......E#.#.#.#.##########
Sample Output
37 5 517 17 9
1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<map> 9 10 #define N 55 11 #define M 15 12 #define mod 6 13 #define mod2 100000000 14 #define ll long long 15 #define maxi(a,b) (a)>(b)? (a) : (b) 16 #define mini(a,b) (a)<(b)? (a) : (b) 17 18 using namespace std; 19 20 int T; 21 int n,m; 22 int c,c1,c2; 23 char s[N][N]; 24 int cc[N][N]; 25 int dirx[]={0,-1,0,1}; 26 int diry[]={-1,0,1,0}; 27 int flag; 28 29 typedef struct 30 { 31 int x; 32 int y; 33 int now; 34 int dir; 35 }PP; 36 37 PP start,end; 38 39 void ini() 40 { 41 int i,j; 42 c=c1=c2=0; 43 //memset(cc,0,sizeof(cc)); 44 scanf("%d%d",&m,&n); 45 for(int i=0;i<n;i++){ 46 scanf("%s",s[i]); 47 } 48 for(i=0;i<n;i++){ 49 for(j=0;j<m;j++){ 50 cc[i][j]=10000000; 51 if(s[i][j]==‘S‘){ 52 start.x=i; 53 start.y=j; 54 start.now=1; 55 } 56 if(s[i][j]==‘E‘){ 57 end.x=i; 58 end.y=j; 59 } 60 } 61 } 62 } 63 64 void solve() 65 { 66 67 } 68 69 int isok(PP o,PP pre) 70 { 71 if( o.x>=0 && o.x<n && o.y>=0 && o.y<m && s[o.x][o.y]!=‘#‘ && cc[o.x][o.y]>pre.now+1) 72 { 73 o.now=pre.now+1; 74 cc[o.x][o.y]=o.now; 75 return o.now; 76 } 77 return 0; 78 } 79 80 void bfs() 81 { 82 int i; 83 PP te,next; 84 queue<PP> q; 85 start.now=1; 86 q.push(start); 87 while(q.size()>0){ 88 te=q.front(); 89 q.pop(); 90 // printf(" %d %d %d\n",te.x,te.y,te.now); 91 for(i=0;i<4;i++){ 92 next.x=te.x+dirx[i]; 93 next.y=te.y+diry[i]; 94 if(isok(next,te)!=0){ 95 next.now=te.now+1; 96 q.push(next); 97 } 98 } 99 }100 c=cc[end.x][end.y];101 }102 103 int ok(PP o)104 {105 if( o.x>=0 && o.x<n && o.y>=0 && o.y<m && s[o.x][o.y]!=‘#‘ )106 {107 return 1;108 }109 return 0;110 }111 112 void dfs1(PP te)113 {114 //flag=0;115 // printf(" %d %d %d %d\n",te.x,te.y,te.dir,te.now);116 int i;117 PP next;118 if(te.x==start.x && te.y==start.y){119 for(i=0;i<4;i++){120 next.x=te.x+dirx[i];121 next.y=te.y+diry[i];122 if(ok(next)!=0){123 next.now=te.now+1;124 next.dir=i;125 dfs1(next);126 }127 }128 }129 130 if(te.x==end.x && te.y==end.y){131 c1=te.now;132 flag=1;133 return;134 }135 136 if(flag==1) return;137 // for(int k=t;k<4;k++){138 i=te.dir-1;139 if(i<0) i+=4;140 next.x=te.x+dirx[i];141 next.y=te.y+diry[i];142 if(ok(next)!=0){143 next.now=te.now+1;144 next.dir=i;145 dfs1(next);146 }147 148 if(flag==1) return;149 150 i=te.dir;151 // if(i<0) i+=4;152 next.x=te.x+dirx[i];153 next.y=te.y+diry[i];154 if(ok(next)!=0){155 next.now=te.now+1;156 next.dir=i;157 dfs1(next);158 }159 160 if(flag==1) return;161 i=te.dir+1;162 if(i>=4) i-=4;163 next.x=te.x+dirx[i];164 next.y=te.y+diry[i];165 if(ok(next)!=0){166 next.now=te.now+1;167 next.dir=i;168 dfs1(next);169 }170 171 if(flag==1) return;172 i=te.dir+2;173 if(i>=4) i-=4;174 next.x=te.x+dirx[i];175 next.y=te.y+diry[i];176 if(ok(next)!=0){177 next.now=te.now+1;178 next.dir=i;179 dfs1(next);180 }181 182 return;183 // }184 185 }186 187 void dfs2(PP te)188 {189 //flag=0;190 // printf(" %d %d %d %d\n",te.x,te.y,te.dir,te.now);191 int i;192 PP next;193 if(te.x==start.x && te.y==start.y){194 for(i=0;i<4;i++){195 next.x=te.x+dirx[i];196 next.y=te.y+diry[i];197 if(ok(next)!=0){198 next.now=te.now+1;199 next.dir=i;200 dfs2(next);201 }202 }203 }204 205 if(te.x==end.x && te.y==end.y){206 c2=te.now;207 flag=1;208 return;209 }210 211 if(flag==1) return;212 // for(int k=t;k<4;k++){213 i=te.dir+1;214 if(i>=4) i-=4;215 next.x=te.x+dirx[i];216 next.y=te.y+diry[i];217 if(ok(next)!=0){218 next.now=te.now+1;219 next.dir=i;220 dfs2(next);221 }222 223 if(flag==1) return;224 225 i=te.dir;226 // if(i<0) i+=4;227 next.x=te.x+dirx[i];228 next.y=te.y+diry[i];229 if(ok(next)!=0){230 next.now=te.now+1;231 next.dir=i;232 dfs2(next);233 }234 235 if(flag==1) return;236 i=te.dir-1;237 if(i<0) i+=4;238 next.x=te.x+dirx[i];239 next.y=te.y+diry[i];240 if(ok(next)!=0){241 next.now=te.now+1;242 next.dir=i;243 dfs2(next);244 }245 246 if(flag==1) return;247 i=te.dir+2;248 if(i>=4) i-=4;249 next.x=te.x+dirx[i];250 next.y=te.y+diry[i];251 if(ok(next)!=0){252 next.now=te.now+1;253 next.dir=i;254 dfs2(next);255 }256 257 return;258 // }259 }260 261 262 int main()263 {264 //freopen("data.in","r",stdin);265 scanf("%d",&T);266 for(int cnt=1;cnt<=T;cnt++)267 //while(T--)268 //while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)269 {270 ini();271 flag=0;272 dfs1(start);273 flag=0;274 dfs2(start);275 bfs();276 printf("%d %d %d\n",c1,c2,c);277 }278 279 return 0;280 }
poj 3083 dfs，bfs
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