POJ 2251 Dungeon Master-白红宇

POJ 2251 Dungeon Master

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发布时间：2019-06-14

本文共 3602 字，大约阅读时间需要 12 分钟。

Dungeon Master

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on

PKU. Original ID:

64-bit integer IO format: %lld Java class name: Main

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?

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.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form

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 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0

Sample Output

Escaped in 11 minute(s).Trapped!

Source

解题：搜索。。。

1 #include 
        
          2 #include 
         
           3 #include 
          
            4 #include 
           
             5 #include 
            
              6 #include 
             
               7 #include 
              
                8 #include 
               
                 9 #include 
                10 #include 
                 
                  11 #include 
                  
                   12 #define LL long long13 #define INF 0x3f3f3f3f14 #define pii pair
                   
                    15 16 using namespace std;17 const int maxn = 40;18 char table[maxn][maxn][maxn];19 int L,R,C,sx,sy,sz,ex,ey,ez;20 bool vis[maxn][maxn][maxn];21 struct node {22 int x,y,z,step;23 node(int tx = 0,int ty = 0,int tz = 0,int tp = 0) {24 x = tx;25 y = ty;26 z = tz;27 step = tp;28 }29 };30 const int dir[6][3] = {31 { 0,0,-1},//left32 { 0,0,1},//right33 { 0,-1,0},//up34 { 0,1,0},//down35 {-1,0,0},//front36 { 1,0,0}//back37 };38 bool isIn(const node &tmp) {39 int a = (tmp.x >= 0 && tmp.x < L);40 int b = (tmp.y >= 0 && tmp.y < R);41 int c = (tmp.z >= 0 && tmp.z < C);42 return a + b + c == 3;43 }44 queue
                    
                     q;45 int bfs() {46 while(!q.empty()) q.pop();47 q.push(node(sx,sy,sz,0));48 vis[sx][sy][sz] = true;49 while(!q.empty()) {50 node now = q.front();51 q.pop();52 if(table[now.x][now.y][now.z] == 'E') return now.step;53 for(int i = 0; i < 6; ++i) {54 node tmp(now.x+dir[i][0],now.y+dir[i][1],now.z+dir[i][2],now.step+1);55 if(isIn(tmp)&&table[tmp.x][tmp.y][tmp.z] != '#' &&!vis[tmp.x][tmp.y][tmp.z]) {56 vis[tmp.x][tmp.y][tmp.z] = true;57 if(table[tmp.x][tmp.y][tmp.z] == 'E') return tmp.step;58 q.push(tmp);59 }60 }61 }62 return -1;63 }64 int main() {65 while(scanf("%d %d %d",&L,&R,&C),L||R||C) {66 memset(table,'\0',sizeof(table));67 memset(vis,false,sizeof(vis));68 for(int i = 0; i < L; ++i) {69 for(int j = 0; j < R; ++j) {70 scanf("%s",table[i][j]);71 for(int k = 0; k < C; ++k)72 if(table[i][j][k] == 'S') {73 sx = i;74 sy = j;75 sz = k;76 }77 }78 }79 int ans = bfs();80 if(~ans) printf("Escaped in %d minute(s).\n",ans);81 else puts("Trapped!");82 }83 return 0;84 }

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转载于:https://www.cnblogs.com/crackpotisback/p/4236289.html

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