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  • 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;
    }
    
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  • 原文地址:https://www.cnblogs.com/SYCstudio/p/7222768.html
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