532 3Ddfs- Dungeon Master

最新推荐文章于 2025-01-17 20:18:02 发布

tianming_fan

最新推荐文章于 2025-01-17 20:18:02 发布

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CC 4.0 BY-SA版权

文章标签： input include zk c

本文链接：https://blog.youkuaiyun.com/tianming_fan/article/details/6630628

本文介绍了一个三维迷宫逃逸问题的解决方案，利用广度优先搜索(BFS)算法找到从起点到终点的最短路径。通过输入迷宫的具体结构，并标记起点与终点，算法能够有效地遍历所有可达节点并找出最优路径。

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#include <iostream>
#include <cstdio>
#include <cstring>
#define MAX 35

using namespace std;

int q[MAX][MAX][MAX];
int visit[MAX][MAX][MAX];
int sx,sy,sz,ex,ey,ez;
int Q[100000];
int L,R,C;

void input(int x,int y,int z)
{
     for(int i = 1;i <= x;i++)
         for(int j = 1;j <= y;j++)
         {
             char s[31];
             cin>>s;

             for(int k = 1;k <= z;k++)
             {
                 if(s[k-1] != '#'){
                     q[i][j][k] = 1;
                     if(s[k-1] == 'S'){
                         sx = i;sy = j;sz = k;
                     }
                     else if(s[k-1] == 'E'){
                         ex = i;ey = j;ez = k;
                     }
                 }
             }
         }
}

int bfs()
{
//printf("%d %d %d %d %d %d\n",sx,sy,sz,ex,ey,ez);

    int dist[MAX][MAX][MAX];
    int dx[6] = {-1,1,0,0,0,0};
    int dy[6] = {0,0,-1,1,0,0};
    int dz[6] = {0,0,0,0,-1,1};
    int x,y,z;
    int front,rear,d,u;
    dist[sx][sy][sz] = 0;
    front = rear = 0;
    u = sx*10000 + sy*100 + sz;
    visit[sx][sy][sz] = 1;
    Q[rear++] = u;

    while(front < rear)
    {
        u = Q[front++];
        x = u / 10000;
        u = u - u/10000*10000;
        y = u / 100;
        z = u % 100;
      // cout<<"++++++++++++++++\n";
      // cout<<x<<" "<<y<<" "<<z<<endl;
         if(x == ex && y ==ey && z == ez){
       //     cout<<"in"<<endl;
            return dist[x][y][z];
         }
//cout<<" _____________________________________\n";

for(d = 0;d < 6;d++)
{

            int nx = x + dx[d],ny = y + dy[d],nz = z + dz[d];
//cout<<nx<<" "<<ny<<" "<<nz<<endl;
            if(nx >=1 && nx <=L && ny >=1 && ny <= R &&
               nz >=1 && nz <=C && !visit[nx][ny][nz] &&
               q[nx][ny][nz]){

                   int v = nx*10000 + ny*100 + nz;
                   Q[rear++] = v;
                   visit[nx][ny][nz] = 1;
                   dist[nx][ny][nz] = dist[x][y][z]+1;

}
}

}

return 0;

}

int main()
{
    while(cin>>L>>R>>C)
    {
        if(!L && !R && !C)
         break;

memset(q,0,sizeof(q));
memset(visit,0,sizeof(visit));

input(L,R,C);

        int temp = bfs();
        if(temp)
         printf("Escaped in %d minute(s).\n",temp);
        else
         cout<<"Trapped!"<<endl;

}

return 0;
}