POJ-2251.DungeonMaster(三维BFS)

做题时需要注意，爬楼有向上和向下爬之分...

　　本题大意：输入 l, r, c, 分别代表地牢的楼层数和每层地牢的长和宽，地牢由rock and point and source and key组成，你初始在s位置，你只能向身边的四个方向和上下方向移动，问你是否能走出地牢，能的话求出最短路径。

　　本题思路：BFS爆搜就OK，我不喜欢存运动方向，所以用循环判断了，按道理循环较慢...所以根据读者喜好选择...

　　参考代码：

#include <cstdio>
#include <iostream>
#include <queue>
#include <cmath>
#include <cstring>
using namespace std;

struct node {
    int x, y, z, step;
} now, Next, source, key;
const int maxn = 30 + 5;
int l, r, c, ans;
char dungeon[maxn][maxn][maxn];
bool vis[maxn][maxn][maxn];

int bfs() {
    memset(vis, false, sizeof vis);
    source.step = 0;
    vis[source.x][source.y][source.z] = true;
    queue <node> Q;
    Q.push(source);
    while(!Q.empty()) {
        now = Q.front();
        Q.pop();
        if(now.x == key.x && now.y == key.y && now.z == key.z)   return now.step;
        Next.step = now.step + 1;
        for(int i = -1; i <= 1; i ++)
            if(i != 0) {
                Next.x = now.x + i, Next.y = now.y, Next.z = now.z;
                if(Next.x >= 0 && Next.y >= 0 && Next.z >= 0 && Next.x < l && Next.y < r && Next.z < c && !vis[Next.x][Next.y][Next.z] && dungeon[Next.x][Next.y][Next.z] != '#' ) {
                    Next.step = now.step + 1;
                    vis[Next.x][Next.y][Next.z] = true;
                    Q.push(Next);
                }
            }
        for(int dy = -1; dy <= 1; dy ++) {
            for(int dz = -1; dz <= 1; dz ++) {
                if((int)(abs(dy - dz)) == 1) {
                    Next.x = now.x + 0, Next.y = now.y + dy, Next.z = now.z + dz;
                    if(Next.x >= 0 && Next.y >= 0 && Next.z >= 0 && Next.x < l && Next.y < r && Next.z < c && !vis[Next.x][Next.y][Next.z] && dungeon[Next.x][Next.y][Next.z] != '#') {
                        vis[Next.x][Next.y][Next.z] = true;
                        Q.push(Next);
                    }
                }       
            }
        }
    }
    return -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 ++) {
                for(int k = 0; k < c; k ++) {
                    cin >> dungeon[i][j][k];
                    if(dungeon[i][j][k] == 'S') { source.x = i; source.y = j; source.z = k; }
                    if(dungeon[i][j][k] == 'E') { key.x = i; key.y = j; key.z = k; }
                }
            }
        }
        ans = bfs();
        if(ans != -1) printf("Escaped in %d minute(s).\n", ans);
        else printf("Trapped!\n");
    }
    return 0;
}