这次整理了一下广度优先搜索的框架，以后可以拿来直接用了。TjuOj1140是一个三维的迷宫题，在BFS时我增加了一个控制数组，用来对队列的出队进行控制，确保每次出队的结点均为同一步长的结点，个人认为比较适合这种迷宫搜索题。
BFS部分的代码如下：
int BFS ( node S , node T ) { int s = 1; //充当指针作用 memset(n,0,sizeof(n)); n[s] = 1; //初始化当前距离为1的点数为1（即原点） node now = S; visit[now.z][now.y][now.x] = 1; node temp; queue<node> Q; Q.push(now); while( !Q.empty() ){ for(int j = 0; j < n[s] ; j++ ){ //依次弹出所有距离为s的方块，进行四周搜索； now = Q.front(); Q.pop(); for(int i = 0 ; i < 6 ;i++){ //向6个方向探索是否有通路 temp.x = now.x + dx[i]; temp.y = now.y + dy[i]; temp.z = now.z + dz[i]; if( visit[temp.z][temp.y][temp.x] == 0 && inmap(temp) ){ //防止越界访问 temp.ch = maze[temp.z][temp.y][temp.x]; if (temp.ch == "E") return s; if (temp.ch == "."){ //如果探索到通路，将该通路标记为(当前距离+1) ,压入队列； n[s + 1]++; //(当前距离+1)的方块个数++ visit[temp.z][temp.y][temp.x] = 1; Q.push(temp); } } } } s++; } return 0; }
题目比较好理解，从S出发找E（但不一定有解），原文如下：
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.
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.
If it is not possible to escape, print the line
Is an escape possible? If yes, how long will it take?
Input Specification
The input file 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 Specification
Each maze generates one line of output. If it is possible to reach the exit, print a line of the formEscaped 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!
/* * 1140_Dungeon Master.cpp * * Created on: 2018年11月14日 * Author: Jeason */ #include <iostream> #include <stdio.h> #include <cstring> #include <sstream> #include <queue> #define N 70 using namespace std; int levels,rows,columns; char maze[N][N][N]; int visit[N][N][N]; int n[ 100000 ] = {0}; //用于记录每个距离的方块个数； string line; int dx[6] = {1,-1,0,0,0,0}; int dy[6] = {0,0,1,-1,0,0}; int dz[6] = {0,0,0,0,1,-1}; struct node { int x,y,z; string ch; }; bool inmap(node A) { if(A.x < 0 || A.x >= columns) return false; if(A.y < 0 || A.y >= rows) return false; if(A.z < 0 || A.z >= levels) return false; return true; } int BFS(node S,node T) { int s = 1; //充当指针作用 memset(n,0,sizeof(n)); n[s] = 1; //初始化当前距离为1的点数为1（即原点） node now = S; visit[now.z][now.y][now.x] = 1; node temp; queue<node> Q; Q.push(now); while( !Q.empty() ){ for(int j = 0; j < n[s] ; j++ ){ //依次弹出所有距离为s的方块，进行四周搜索； now = Q.front(); Q.pop(); for(int i = 0 ; i < 6 ;i++){ //向6个方向探索是否有通路 temp.x = now.x + dx[i]; temp.y = now.y + dy[i]; temp.z = now.z + dz[i]; if( visit[temp.z][temp.y][temp.x] == 0 && inmap(temp) ){ //防止越界访问 temp.ch = maze[temp.z][temp.y][temp.x]; if (temp.ch == "E") return s; if (temp.ch == "."){ //如果探索到通路，将该通路标记为(当前距离+1) ,压入队列； n[s + 1]++; //(当前距离+1)的方块个数++ visit[temp.z][temp.y][temp.x] = 1; Q.push(temp); } } } } s++; } return 0; } int main() { node S,T; cin >> levels >> rows >> columns; while( (levels != 0)&&(rows != 0)&&(columns != 0) ){ memset(maze, 0, sizeof(maze)); //读入数据； memset(visit, 0, sizeof(visit)); for(int i = 0;i < levels ;i++){ for(int j = 0;j < rows ;j++){ cin >> line; for(int k = 0;k < columns ;k++){ maze[i][j][k] = line[k]; if(line[k] == 'S'){ //找起点 S.z = i; S.x = k; S.y = j; S.ch = "S"; } } } } int minutes = BFS( S, T ); if(minutes != 0) cout << "Escaped in "<< minutes << " minute(s)."<< endl; else cout << "Trapped!" <<endl; // for(int i = 0;i < levels ;i++){ //输出地图 // for(int j = 0;j < rows ;j++){ // for(int k = 0;k < columns ;k++){ // cout << maze[i][j][k]; // } // cout << endl; // } // cout << endl; // } cin >> levels >> rows >> columns; } return 0; } /* Sample Input 3 4 5 S.... .###. .##.. ###.# ##### ##### ##.## ##... ##### ##### #.### ####E 1 3 3 S## #E# ### 0 0 0 Sample Output Escaped in 11 minute(s). Trapped! */