1 条题解

  • 1
    @ 2023-11-11 16:25:12
    #include <iostream>
    #include <cstring>
    #include <queue> 
    using namespace std;
    
    const int N = 110;
    
    struct Node
    {
    	int x, y, z;
    };
    
    char g[N][N][N];
    int dist[N][N][N];
    int L, R, C, x1, y1, z1, x2, y2, z2;
    
    int dx[6] = {-1, 0, 1, 0, 0, 0};
    int dy[6] = {0, 1, 0, -1, 0, 0};
    int dz[6] = {0, 0, 0, 0, -1, 1};
    
    int bfs()
    {
        memset(dist, -1, sizeof dist);
    	queue<Node> q;
    	q.push({x1, y1, z1});
    	dist[x1][y1][z1] = 0;
    	
    	while(q.size())
    	{
    		Node t = q.front();
    		q.pop();
    		
    		if(t.x == x2 && t.y == y2 && t.z == z2) return dist[t.x][t.y][t.z];
    		
    		for (int i = 0; i < 6; i ++)
    		{
    			int a = t.x + dx[i], b = t.y + dy[i], c = t.z + dz[i];
    			if(a < 1 || a > L || b < 1 || b > R || c < 1 || c > C) continue;
    			if(g[a][b][c] == '#' || dist[a][b][c] != -1) continue;
    			
    			q.push({a, b, c});
    			dist[a][b][c] = dist[t.x][t.y][t.z] + 1;
    		}
    	}
    	
    	return -1;
    }
    
    int main()
    {
        while(cin >> L >> R >> C, L || R || C)
        {
            for (int i = 1; i <= L; i ++)
        		for (int j = 1; j <= R; j ++)
        			for (int k = 1; k <= C; k ++)
        			{
        				cin >> g[i][j][k];
        				if(g[i][j][k] == 'S') x1 = i, y1 = j, z1 = k;
        				if(g[i][j][k] == 'E') x2 = i, y2 = j, z2 = k;
        			}
        	
        	if(bfs() == -1) cout << "Trapped!" << endl;
        	else cout << "Escaped in " << dist[x2][y2][z2] << " minute(s)." << endl;		
        }
    	
    	return 0;		
    }
    


    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<string>
    #include<cstdlib>
    #include<queue>
    #include<vector>
    #define INF 0x3f3f3f3f
    #define PI acos(-1.0)
    #define N 101
    #define MOD 2520
    #define E 1e-12
    using namespace std;
    int X,Y,Z;
    char a[N][N][N];
    bool vis[N][N][N];
    int dir[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
    struct node
    {
        int x;
        int y;
        int z;
        int step;
    }q[N*10000];
    void bfs(int sx,int sy,int sz,int ex,int ey,int ez)
    {
        int head=1,tail=1;
        bool flag=true;
        memset(vis,0,sizeof(vis));
     
        vis[sx][sy][sz]=1;
        q[tail].x=sx;
        q[tail].y=sy;
        q[tail].z=sz;
        q[tail].step=0;
        tail++;
     
        while(head<tail)
        {
            int x=q[head].x;
            int y=q[head].y;
            int z=q[head].z;
            int step=q[head].step;
            if(x==ex&&y==ey&&z==ez)
            {
                flag=false;
                printf("Escaped in %d minute(s).\n",step);
                break;
            }
            for(int i=0;i<6;i++)
            {
                int nx=x+dir[i][0];
                int ny=y+dir[i][1];
                int nz=z+dir[i][2];
                if(0<=nx&&nx<X&&0<=ny&&ny<Y&&0<=nz&&nz<Z&&a[nx][ny][nz]=='.'&&vis[nx][ny][nz]==0)
                {
                    vis[nx][ny][nz]=1;
                    q[tail].x=nx;
                    q[tail].y=ny;
                    q[tail].z=nz;
                    q[tail].step=step+1;
                    tail++;
                }
            }
            head++;
        }
        if(flag)
            printf("Trapped!\n");
     
    }
    int main()
    {
        int sx,sy,sz,ex,ey,ez;
        while(scanf("%d%d%d",&X,&Y,&Z)!=EOF&&X&&Y&&Z)
        {
            for(int i=0;i<X;i++)
                for(int j=0;j<Y;j++)
                    scanf("%s",a[i][j]);
            for(int i=0;i<X;i++)
                for(int j=0;j<Y;j++)
                    for(int k=0;k<Z;k++)
                    {
                        if(a[i][j][k]=='S')
                        {
                            sx=i;
                            sy=j;
                            sz=k;
                        }
                        if(a[i][j][k]=='E')
                        {
                            ex=i;
                            ey=j;
                            ez=k;
                            a[i][j][k]='.';
                        }
                    }
            bfs(sx,sy,sz,ex,ey,ez);
        }
        return 0;
    }
    
    • 1

    信息

    ID
    424
    时间
    1000ms
    内存
    256MiB
    难度
    9
    标签
    递交数
    11
    已通过
    5
    上传者