zoukankan      html  css  js  c++  java
  • 迷宫问题-广度优先搜索

    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 5
    S....
    .###.
    .##..
    ###.#

    #####
    #####
    ##.##
    ##...

    #####
    #####
    #.###
    ####E

    1 3 3
    S##
    #E#
    ###

    0 0 0


    Sample Output

    Escaped in 11 minute(s).

    Trapped!


    题解:最初我试了一下动态规划,结果失败,为什么呢?因为动态规划先做什么后做什么必须明确,而迷宫问题却不知道先算谁后算谁.还是广度优先快.

    实现队列非常简单,一定要手动背写下来.

    #include<iostream>
    #include<math.h>
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<time.h>
    #include<stdlib.h>
    using namespace std; 
    int a[31][31][31];
    char map[31][31][31];
    int xsize, ysize, zsize;
    struct Point{
    	int x, y, z;
    };
    struct queue{
    	Point a[10000];
    	int head, rear;
    	void init(){
    		head = rear = 0;
    	}
    	void enq(int x,int y ,int z){
    		Point&p = a[rear];
    		p.x = x;
    		p.y = y;
    		p.z = z;
    		rear++;
    		rear %= 10000;
    	}
    	Point& deq(){
    		int temp = head;
    		head++;
    		head %= 10000;
    		return a[temp];
    	}
    	bool isEmpty(){
    		return rear == head;
    	}
    };
    queue q;
    int sx, sy, sz;
    int ans=-1;
    void check(int x, int y, int z,int father){
    	if (x < 0 || y < 0 || z < 0)return;
    	if (x >= xsize || y >= ysize || z >= zsize)return;
    	if (~a[x][y][z])return;
    	if (map[x][y][z] == '#')return;
    	if (map[x][y][z] == 'E'){
    		ans = father + 1;
    		return;
    	}
    	a[x][y][z] = father + 1;
    	q.enq(x, y, z);
    }
    void go(){
    	memset(a, -1, sizeof(a));
    	ans = -1;
    	q.init();
    	q.enq(sx, sy, sz);
    	a[sx][sy][sz] = 0;
    	while (!q.isEmpty()){
    		Point &p = q.deq();
    		int&x = p.x;
    		int &y = p.y;
    		int &z = p.z; 
    		int &f = a[x][y][z];
    		check(x + 1, y, z,f);
    		check(x - 1, y, z,f);
    		check(x, y + 1, z,f);
    		check(x, y - 1, z,f);
    		check(x, y, z + 1,f);
    		check(x, y, z - 1,f);
    		if (~ans)return;
    	}
    }
    int main(){ 
    	while (cin >> xsize >> ysize >> zsize && (xsize || ysize || zsize)){
    		int i, j, k; 
    		for (i = 0; i < xsize; i++){
    			for (j = 0; j < ysize; j++){
    				for (k = 0; k < zsize; k++){
    					cin >> map[i][j][k];
    					if (map[i][j][k] == 'S'){
    						sx = i;
    						sy = j;
    						sz = k;
    						a[i][j][k] = -1;
    					} 
    				}
    			}
    		} 
    		go();
    		if (ans==-1 ) cout << "Trapped!" << endl;
    		else cout << "Escaped in " << ans << " minute(s)." << endl;
    	}
    	return 0;
    }


  • 相关阅读:
    Block代码块中使用局部变量注意点
    This application's application-identifier entitlement does not match that of the installed application. These values must match for an upgrade to be allowed.
    iOS实现渐变颜色
    linker command failed with exit code 1 (use -v to see invocation)
    关于SVN的405错误Server sent unexpected return value (405 Method Not Allowed)的解决办法
    提交SVN Working copy locked解决
    FSCalendar使用和注意事项
    c/c++左值和右值
    mysql使用笔记(一)
    解决 ssh 登录到ubuntu server 慢的问题
  • 原文地址:https://www.cnblogs.com/weiyinfu/p/5013886.html
Copyright © 2011-2022 走看看