POJ 2251 Dungeon Master【三维迷宫裸BFS练习】
来源:互联网 发布:魔兽世界数据库2.43 编辑:程序博客网 时间:2024/05/16 07:39
链接:
http://poj.org/problem?id=2251
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22009#problem/E
Dungeon Master
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14103 Accepted: 5477
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 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0
Sample Output
Escaped in 11 minute(s).Trapped!
Source
Ulm Local 1997
code:
#include<stdio.h>#include<string.h>#include<queue>#include<iostream>using namespace std;const int maxn = 40;int map[maxn][maxn][maxn];int vis[maxn][maxn][maxn];char str[maxn];int L,R,C;int flag;int step;int dir[6][3] = {1,0,0, 0,0,1, 0,1,0, 0,0,-1, 0,-1,0, -1,0,0};struct Point{ int x,y,z; int step;}st,en;void bfs(int x, int y, int z){ queue<Point> q; while(!q.empty()) q.pop(); //清空队列 Point now,next; now.x = x; now.y = y; now.z = z; now.step = 0; q.push(now); //起点入队 memset(vis, 0, sizeof(vis)); vis[x][y][z] = 1; while(!q.empty()) { now = q.front(); q.pop(); if(now.x == en.x && now.y == en.y && now.z == en.z) { flag = 1; //标记找到 en.step = now.step; // return; } for(int i = 0; i < 6; i++) { next.x = now.x+dir[i][0]; next.y = now.y+dir[i][1]; next.z = now.z+dir[i][2]; if(map[next.x][next.y][next.z] && !vis[next.x][next.y][next.z]) { vis[next.x][next.y][next.z] = 1; next.step = now.step+1; q.push(next); if(next.x == en.x && next.y == en.y && next.z == en.z) { flag = 1; en.step = next.step; return; } } } } return;}int main(){ while(scanf("%d%d%d", &L,&R,&C) != EOF) { if(L == 0 && R == 0 && C == 0) break; memset(map,0,sizeof(map)); gets(str); char c; for(int i = 1; i <= L; i++) { for(int j = 1; j <= R; j++) { for(int k = 1; k <= C; k++) { scanf("%c", &c); if(c == 'S') { st.x = i; st.y = j; st.z = k; } if(c == 'E') { en.x = i; en.y = j; en.z = k; } if(c != '#') map[i][j][k] = 1; //标记可以走标记为 1,避免了判断边界问题 } gets(str); } gets(str); } flag = 0; //初始化 bfs(st.x, st.y, st.z); if(flag == 0) printf("Trapped!\n"); else printf("Escaped in %d minute(s).\n", en.step); } return 0;}
- POJ 2251 Dungeon Master【三维迷宫裸BFS练习】
- POJ 2251 Dungeon Master (三维迷宫 BFS)
- POJ NO.2251 Dungeon Master(BFS,三维迷宫)
- POJ-2251-Dungeon Master【BFS】【三维迷宫问题】
- POJ 2251-Dungeon Master(BFS-三维迷宫)
- POJ 2251 Dungeon Master (三维BFS)
- POJ 2251:Dungeon Master(三维BFS)
- POJ 2251 Dungeon Master(三维BFS)
- poj 2251 Dungeon Master 三维bfs
- poj 2251 Dungeon Master(三维BFS)(中等)
- poj 2251 Dungeon Master(BFS三维)
- POJ 2251 Dungeon Master(三维bfs)
- POJ 2251 Dungeon Master(三维bfs)
- <三维BFS搜索> POJ 2251 Dungeon Master
- POJ 2251:Dungeon Master(三维BFS)
- POJ 2251 Dungeon Master(三维BFS)
- POJ 2251 Dungeon Master 三维BFS
- POJ 2251 Dungeon Master(三维bfs)
- hdu1069 Monkey and Banana
- C++ vector容器类型
- Linux 设备驱动开发详解之20章usb主机与设备驱动
- java.io.FilenameFilter文件名过滤器总结
- BNUOJ Scarecrow 题解
- POJ 2251 Dungeon Master【三维迷宫裸BFS练习】
- “Linux文件的合并、排序和分割” 之 split 命令
- 2345导航高价卖身-网络是否又有新变革?
- Linux中断处理之系统调用
- 删除数据库中所有的表
- void assert( int expression );
- 大话设计模式之桥接模式
- android MotionEvent中getX()和getRawX()的区别
- Maven3:用eclipse插件创建一个web project