zoukankan      html  css  js  c++  java
  • POJ 2251 Dungeon Master /UVA 532 Dungeon Master / ZOJ 1940 Dungeon Master(广度优先搜索)

    #POJ 2251 Dungeon Master /UVA 532 Dungeon Master / ZOJ 1940 Dungeon Master(广度优先搜索)

    ###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? ###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! ###Http POJ:https://vjudge.net/problem/POJ-2251 ###Source 广度优先搜索 ##题目大意 在一个三维的迷宫中从起点走到终点 ##解决思路 广度优先搜索,向六个方向延伸 ##代码

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    
    const int maxN=35;
    const int inf=2147483647;
    const int F1[10]={0,1,-1,0,0,0,0};
    const int F2[10]={0,0,0,1,-1,0,0};
    const int F3[10]={0,0,0,0,0,1,-1};
    
    class Position
    {
    public:
        int x,y,z;
        int step;
    };
    
    int a,b,c;
    int Map[maxN][maxN][maxN];
    queue<Position> Q;
    bool vis[maxN][maxN][maxN];
    
    int main()
    {
        while (cin>>a>>b>>c)
        {
            if ((a==0)&&(b==0)&&(c==0))
                break;
            memset(Map,-1,sizeof(Map));
            memset(vis,0,sizeof(vis));
            while (!Q.empty())
                Q.pop();
            int x0,y0,z0,x1,y1,z1;
            char str[maxN];
            for (int i=1;i<=a;i++)
                for (int j=1;j<=b;j++)
                {
                    cin>>str;
                    for (int k=0;k<c;k++)
                    {
                        if (str[k]!='#')
                            Map[i][j][k+1]=1;
                        if (str[k]=='S')
                        {
                            x0=i;
                            y0=j;
                            z0=k+1;
                        }
                        else
                        if (str[k]=='E')
                        {
                            x1=i;
                            y1=j;
                            z1=k+1;
                        }
                    }
                }
            /*for (int i=1;i<=a;i++)
            {
                for (int j=1;j<=b;j++)
                {
                    for (int k=1;k<=c;k++)
                        cout<<Map[i][j][k]<<' ';
                    cout<<endl;
                }
                cout<<endl;
            }*/
            Q.push((Position){x0,y0,z0,0});
            vis[x0][y0][z0]=1;
            bool is_get=0;
            do
            {
                Position u=Q.front();
                Q.pop();
                //cout<<u.x<<' '<<u.y<<' '<<u.z<<' '<<u.step<<endl;
                if ((u.x==x1)&&(u.y==y1)&&(u.z==z1))
                {
                    is_get=1;
                    //out<<u.step<<endl;
                    printf("Escaped in %d minute(s).
    ",u.step);
                    break;
                }
                for (int i=1;i<=6;i++)
                    {
                        int x2=u.x+F1[i];
                        int y2=u.y+F2[i];
                        int z2=u.z+F3[i];
                        if ((Map[x2][y2][z2]!=-1)&&(vis[x2][y2][z2]==0))
                        {
                            //cout<<"("<<u.x<<","<<u.y<<','<<u.z<<")->("<<x2<<","<<y2<<","<<z2<<")"<<endl;
                            Q.push((Position){x2,y2,z2,u.step+1});
                            vis[x2][y2][z2]=1;
                        }
                    }
            }
            while (!Q.empty());
            if (is_get==0)
                cout<<"Trapped!"<<endl;
        }
        return 0;
    }
    
  • 相关阅读:
    哈希表详解
    简单字典实现(KV问题)
    【数据结构】——搜索二叉树的插入,查找和删除(递归&非递归)
    【数据结构】——堆及其应用
    初学者的迷茫
    【数据结构】顺序表和链表
    程序运行机理
    回调函数到底是怎么一回事呢?
    C语言中函数可变参数解析
    《剑指offer》数组中出现次数超过数组长度一半的数字
  • 原文地址:https://www.cnblogs.com/SYCstudio/p/7222768.html
Copyright © 2011-2022 走看看