• POJ:Dungeon Master(BFS模板题)


    原题链接

    思路:

    正常的思路,只不过是将二维BFS换成三维的,也算是个模板题吧(PS:DFS超容易超时)

    #include<iostream>
    #include<queue>
    #include<cstring>
    using namespace std;
    char chart[40][40][40];
    
    int then[6][3] = { {-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,1},{0,0,-1} };
    int l, r, c;
    int el, er, ec, sl, sr, sc;
    int ans;
    struct node
    {
        int step;
        int l;
        int r;
        int c;
    };
    bool judge(int x, int y, int z)
    {
        if (x >= 0 && x < l && y >= 0 && y < r && z >= 0 && z < c && chart[x][y][z] != '#')
            return true;
        else
            return false;
    }
    void bfs()
    {
        int flag = 0;
        queue<node> q;
        node t;
        t.step = 0;
        t.l = sl;
        t.r = sr;
        t.c = sc;
        q.push(t);
        chart[sl][sr][sc] = '#';
        while (!q.empty())
        {
            int x, y, z;
            x = q.front().l;
            y = q.front().r;
            z = q.front().c;
            for (int i = 0; i < 6; i++)
            {
                int xx = x + then[i][0];
                int yy = y + then[i][1];
                int zz = z + then[i][2];
                if (judge(xx, yy, zz))
                {
                    if (xx == el && yy == er && zz == ec)//在找到的时候就退出,如果在走到的时候再退出的话,会t
                    {
                        ans = q.front().step + 1;
                        return;
                    }
                    node temp;
                    temp.step = q.front().step + 1;
                    temp.l = xx;
                    temp.r = yy;
                    temp.c = zz;
                    chart[xx][yy][zz] = '#';
                    q.push(temp);
                }
            }
            q.pop();
        }
        ans = -1;
    }
    int main()
    {
        while (scanf("%d%d%d", &l, &r, &c))
        {
            if (l == 0 && r == 0 && c == 0)
                break;
    
            for (int i = 0; i < l; i++)
            {
                for (int j = 0; j < r; j++)
                {
                    scanf("%s", chart[i][j]);
                    for (int k = 0; k < c; k++)
                    {
                        if (chart[i][j][k] == 'S')
                        {
                            sl = i;
                            sr = j;
                            sc = k;
                        }
                        if (chart[i][j][k] == 'E')
                        {
                            el = i;
                            er = j;
                            ec = k;
                        }
                    }
                }
            }
            bfs();
            if (ans == -1)
                printf("Trapped!
    ");
            else
                printf("Escaped in %d minute(s).
    ", ans);
        }
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/RioTian/p/12842502.html
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