深搜,只不过是三维的. #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; int l,r,c; int sx,sy,sz; int ex,ey,ez; ; ][][]; ][][]= {}; ; void dfs(int x,int y,int z,int step) { if(x==ex&&y==ey&…

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Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16748   Accepted: 6522 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…

题目链接:http://poj.org/problem?id=2251 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37296   Accepted: 14266 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit…

1棋盘问题 在一个给定形状的棋盘(形状可能是不规则的)上面摆放棋子,棋子没有区别.要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列,请编程求解对于给定形状和大小的棋盘,摆放k个棋子的所有可行的摆放方案C. Input 输入含有多组测试数据. 每组数据的第一行是两个正整数,n k,用一个空格隔开,表示了将在一个n*n的矩阵内描述棋盘,以及摆放棋子的数目. n <= 8 , k <= n 当为-1 -1时表示输入结束. 随后的n行描述了棋盘的形状:每行有n个字符,其中 # 表示棋盘区域,…

传送门 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28416   Accepted: 11109 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 f…

http://poj.org/problem?id=2251 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18773   Accepted: 7285 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes…

Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20995 Accepted: 8150 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 wit…

POJ 2251 题目大意: 给出一三维空间的地牢,要求求出由字符'S'到字符'E'的最短路径,移动方向可以是上,下,左,右,前,后,六个方向,每移动一次就耗费一分钟,要求输出最快的走出时间.不同L层的地图,相同RC坐标处是相连通的.(.可走,#为墙) 解题思路:从起点开始分别往6个方向进行BFS(即入队),并记录步数,直至队为空.若一直找不到,则困住. /* POJ 2251 Dungeon Master --- 三维BFS(用BFS求最短路) */ #include <cstdio> #i…

    学习点: scanf可以自动过滤空行 搜索时要先判断是否越界(L R C),再判断其他条件是否满足 bfs搜索时可以在入口处(push时)判断是否达到目标,也可以在出口处(pop时)   #include<cstdio> #include<cstring> #include<iostream> #include<string> #include<queue> using namespace std; ; , , , , , }; , ,…