Dungeon Master bfs
Description
You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.
Is an escape possible? If yes, how long will it take?
Is an escape possible? If yes, how long will it take?
Input
The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.
Output
Each maze generates one line of output. If it is possible to reach the exit, print a line of the form
where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line
Escaped in x minute(s).
where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line
Trapped!
Sample Input
3 4 5 S.... .###. .##.. ###.# ##### ##### ##.## ##... ##### ##### #.### ####E 1 3 3 S## #E# ### 0 0 0
Sample Output
Escaped in 11 minute(s). Trapped!
queue用法
queue 模板类的定义在<queue>头文件中。
queue 模板类需要两个模板参数，一个是元素类型，一个容器类
型，元素类型是必要的，容器类型是可选的，默认为deque 类型。
定义queue 对象的示例代码如下：
queue<int> q1;
queue<double> q2;
queue 的基本操作有：
入队，如例：q.push(x); 将x 接到队列的末端。
出队，如例：q.pop(); 弹出队列的第一个元素，注意，并不会返回被弹出元素的值。
访问队首元素，如例：q.front()，即最早被压入队列的元素。
访问队尾元素，如例：q.back()，即最后被压入队列的元素。
判断队列空，如例：q.empty()，当队列空时，返回true。
访问队列中的元素个数，如例：q.size()
1 #include<stdio.h> 2 #include<string.h> 3 #include<queue> 4 using namespace std; 5 6 int main() 7 { 8 queue<int> q; 9 q.push(1); 10 q.push(3); 11 q.push(5); 12 printf("%d\n",q.size()); 13 while(!q.empty()) 14 { 15 int a=q.front(); 16 q.pop(); 17 printf("%d ",a); 18 } 19 printf("\n"); 20 return 0; 21 }
my daima
1 #include<stdio.h> 2 #include<string.h> 3 #include<queue> 4 using namespace std; 5 6 const int INF=9999999; 7 int l,r,c,sl,sr,sc,el,er,ec,d[35][35][35]; 8 char maps[35][35][35]; 9 10 struct Node 11 { 12 int l; 13 int r; 14 int c; 15 }; 16 17 queue<Node> q; 18 19 int dr[6]={0,0,1,-1,0,0},dc[6]={0,0,0,0,1,-1},dl[6]={1,-1,0,0,0,0}; 20 21 int bfs() 22 { 23 int i,j,k; 24 for(k=1;k<=l;k++) 25 for(i=1;i<=r;i++) 26 for(j=1;j<=c;j++) 27 d[k][i][j]=INF; 28 Node s; 29 s.l=sl,s.r=sr,s.c=sc; 30 q.push(s); 31 d[sl][sr][sc]=0; 32 33 while(q.size()) 34 { 35 Node now=q.front(); 36 q.pop(); 37 Node nex; 38 39 if(now.l==el && now.r==er && now.c==ec) 40 break; 41 42 for(i=0;i<6;i++) 43 { 44 nex.l=now.l+dl[i],nex.r=now.r+dr[i],nex.c=now.c+dc[i]; 45 46 if(1<=nex.l && nex.l<=l && 1<=nex.r && nex.r<=r && 1<=nex.c && nex.c<=c && maps[nex.l][nex.r][nex.c]!='#' && d[nex.l][nex.r][nex.c]==INF) 47 { 48 q.push(nex); 49 d[nex.l][nex.r][nex.c]=d[now.l][now.r][now.c]+1; 50 } 51 } 52 } 53 return d[el][er][ec]; 54 } 55 56 int main() 57 { 58 int i,j,k; 59 while(scanf("%d %d %d",&l,&r,&c)!=EOF) 60 { 61 getchar(); 62 if(l==0 && r==0 && c==0) 63 break; 64 for(k=1;k<=l;k++) 65 { 66 for(i=1;i<=r;i++) 67 { 68 for(j=1;j<=c;j++) 69 { 70 scanf("%c",&maps[k][i][j]); 71 if(maps[k][i][j]=='S') 72 sl=k,sr=i,sc=j; 73 if(maps[k][i][j]=='E') 74 el=k,er=i,ec=j; 75 } 76 getchar(); 77 } 78 getchar(); 79 } 80 81 int ans=bfs(); 82 if(ans==INF) 83 printf("Trapped!\n"); 84 else 85 printf("Escaped in %d minute(s).\n",ans); 86 87 88 /*printf("%d %d %d\n%d %d %d\n",sl,sr,sc,el,er,ec); 89 for(k=1;k<=l;k++) 90 { 91 for(i=1;i<=r;i++) 92 { 93 for(j=1;j<=c;j++) 94 { 95 printf("%7d ",d[k][i][j]); 96 } 97 printf("\n"); 98 } 99 printf("\n"); 100 }*/ 101 102 } 103 return 0; 104 }
dsdm
#include <iostream> #include <cstring> #include <string> #include <cstdlib> #include <queue> using namespace std; int l, r, c, res; int sx, sy, sz, ex, ey, ez; string m[31][31]; bool vis[31][31][31], esp; struct Node {int x, y, z, rs;}; int dir[6][3] = {-1,0,0,1,0,0,0,-1,0,0,1,0,0,0,-1,0,0,1}; queue<Node> q; void bfs() { Node s; s.x = sx, s.y = sy, s.z = sz, s.rs = 0; while(!q.empty()) q.pop(); q.push(s); while(!q.empty()) { Node now = q.front(); q.pop(); Node tmp; for (int i = 0; i < 6; ++i) { int x = now.x+dir[i][0]; int y = now.y+dir[i][1]; int z = now.z+dir[i][2]; if (x >= l || x < 0 || y >= r || y < 0 || z >= c || z < 0) continue; if (vis[z][x][y]) continue; if (m[z][x][y] =='#') continue; vis[z][x][y] = true; tmp.x = x; tmp.y = y; tmp.z = z; tmp.rs = now.rs +1; q.push(tmp); if (x == ex && y == ey && z == ez) { esp = true; res = tmp.rs; return ; } } } return ; } int main() { while(cin >> c >> l >> r) { if (!l && !r && !c) break; for (int i = 0; i < c; ++i) { for (int j = 0; j < l; ++j) { cin >> m[i][j]; for (int k = 0; k < m[i][j].size(); ++k) { if (m[i][j][k] == 'S') { sx = j; sy = k; sz = i; } else if (m[i][j][k] == 'E') { ex = j; ey = k; ez = i; } } } } memset(vis, 0, sizeof(vis)); vis[sz][sx][sy] = true; esp = false; bfs(); if (esp) cout << "Escaped in "<<res<< " minute(s)." << endl; else cout << "Trapped!" << endl; } return 0; }
