POJ 2251 Dungeon Master --- 三维BFS(用BFS求最短路)

　　POJ 2251

　　题目大意： 给出一三维空间的地牢,要求求出由字符‘S‘到字符‘E‘的最短路径，移动方向可以是上，下，左，右，前，后，六个方向，每移动一次就耗费一分钟，要求输出最快的走出时间。不同L层的地图，相同RC坐标处是相连通的。(.可走，#为墙)

　　解题思路:从起点开始分别往6个方向进行BFS(即入队),并记录步数,直至队为空.若一直找不到,则困住。

/* POJ 2251 Dungeon Master --- 三维BFS(用BFS求最短路) */
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

int n, m, o; //n行m列,o为层
char mapp[31][31][31];
bool visit[31][31][31]; //标记访问状态

/* 构造队列元素 (点位置+步数)*/
struct Point{
    int x, y, z;
    int step;
    Point(){}
    Point(int a, int b, int c, int d) :x(a), y(b), z(c), step(d){}
};

/* 判断点是否满足访问条件 */
inline bool judge(Point tmp){
    if (mapp[tmp.x][tmp.y][tmp.z] == ‘#‘ || visit[tmp.x][tmp.y][tmp.z]
        || tmp.x < 0 || tmp.x >= o || tmp.y < 0 || tmp.y >= n
        || tmp.z < 0 || tmp.z >= m
        ){
        return 0;
    }
    return 1;
}

/* x0为深度 y0,z0为同一层坐标 */
void bfs(int x0, int y0,int z0){
    queue<Point> q;
    bool found = 0;
    visit[x0][y0][z0] = 1;  //起点标记已访问
    q.push(Point(x0, y0, z0, 0));
    while (!q.empty()){
        Point tmp = q.front(); q.pop();
        if (mapp[tmp.x][tmp.y][tmp.z] == ‘E‘){
            //找到终点，搜索结束，输出最短路
            printf("Escaped in %d minute(s).\n", tmp.step);
            found = 1;
            break;
        }
        else{
            Point tmp2;
            ++tmp.step; //步数必定增加

            //往上一层
            tmp2 = tmp;
            tmp2.x = tmp.x - 1;
            if (judge(tmp2)){
                visit[tmp2.x][tmp2.y][tmp2.z] = 1;
                q.push(tmp2);
            }

            //往下一层
            tmp2 = tmp;
            tmp2.x = tmp.x + 1;
            if (judge(tmp2)){
                visit[tmp2.x][tmp2.y][tmp2.z] = 1;
                q.push(tmp2);
            }

            //以下四种为同一层

            //往上一行
            tmp2 = tmp;
            tmp2.y = tmp.y - 1;
            if (judge(tmp2)){
                visit[tmp2.x][tmp2.y][tmp2.z] = 1;
                q.push(tmp2);
            }

            //往下一行
            tmp2 = tmp;
            tmp2.y = tmp.y + 1;
            if (judge(tmp2)){
                visit[tmp2.x][tmp2.y][tmp2.z] = 1;
                q.push(tmp2);
            }

            //往左一列
            tmp2 = tmp;
            tmp2.z = tmp.z - 1;
            if (judge(tmp2)){
                visit[tmp2.x][tmp2.y][tmp2.z] = 1;
                q.push(tmp2);
            }

            //往右一列
            tmp2 = tmp;
            tmp2.z = tmp.z + 1;
            if (judge(tmp2)){
                visit[tmp2.x][tmp2.y][tmp2.z] = 1;
                q.push(tmp2);
            }

        }//else
    }//while
    if (!found){
        printf("Trapped!\n");
    }
}

int main()
{
#ifdef _LOCAL
    freopen("D:\\input.txt", "r", stdin);
#endif

    int sx, sy, sz;
    while (scanf("%d%d%d", &o, &n, &m) == 3 && (m + n + o)){
        memset(visit, 0, sizeof visit);
        for (int i = 0; i < o; ++i){
            for (int j = 0; j < n; ++j){
                scanf("%s", mapp[i][j]);
                for (int k = 0; k < m; ++k){
                    if (mapp[i][j][k] == ‘S‘){
                        sx = i;
                        sy = j;
                        sz = k;
                    }
                }//for(k)
            }//for(j) n行
        }//for(i) o层
        bfs(sx, sy, sz); //sx是深度
    }

    return 0;
}