题目描述

bfs 

(一)WE
if(nx<1||nx>r||ny<1||ny>c||nz<1||nz>l||g[nx][ny][nz]=='#'||st[nx][ny][nz]){continue;}


(二)
bool check(int nx, int ny, int nz) 
{
    return nx<1||nx>r||ny<1||ny>c||nz<1||nz>l;
}
if(check(nx,ny,nz)||g[nx][ny][nz]=='#'||st[nx][ny][nz]){continue;}

算法1

C++ 代码

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int INF=0x3f3f3f;
const int N=110;
int l,r,c;
char g[N][N][N];
int dist[N][N][N];
bool st[N][N][N];

struct position{
    int x,y,z;
};

int dx[6]={-1,1,0,0,0,0},dy[6]={0,0,1,-1,0,0},dz[6]={0,0,0,0,1,-1};

bool check(int nx, int ny, int nz) //判断是否在地图内
{
    return nx<1||nx>r||ny<1||ny>c||nz<1||nz>l;
}

void bfs(position start,position end){
    memset(dist,-1,sizeof(dist));
    memset(st,false,sizeof(st));
    queue<position> q;
    dist[start.x][start.y][start.z]=0;
    st[start.x][start.y][start.z]=true;

    q.push(start);

    while(q.size()){
        position t=q.front();q.pop();
        int a=t.x,b=t.y,c=t.z;

        for(int i=0;i<6;i++){
            int nx=a+dx[i],ny=b+dy[i],nz=c+dz[i];

            // if(nx<1||nx>r||ny<1||ny>c||nz<1||nz>l||g[nx][ny][nz]=='#'||st[nx][ny][nz]){continue;}
            //为什么上面的不可以，但是下面的可以？？
            if(check(nx,ny,nz)||g[nx][ny][nz]=='#'||st[nx][ny][nz]){continue;}
            if(dist[nx][ny][nz]==-1){
                dist[nx][ny][nz]=dist[a][b][c]+1;
                st[nx][ny][nz]=true;
                q.push({nx,ny,nz});
            }

        }
    }

}

int main(){
    while(cin>>l>>r>>c){
        if(l==0&&r==0&&c==0){
            break;
        }
        position start,end;
        for(int k=1;k<=l;k++){
            for(int i=1;i<=r;i++){
                for(int j=1;j<=c;j++){
                    cin>>g[i][j][k];
                    if(g[i][j][k]=='S'){
                        start={i,j,k};
                    }
                    if(g[i][j][k]=='E'){
                        end={i,j,k};    
                    }
                }
            }
        }

        bfs(start,end);

        if(dist[end.x][end.y][end.z]==-1){
            cout<<"Trapped!"<<endl;
        }else{
            cout<<"Escaped in "<<dist[end.x][end.y][end.z]<<" minute(s)."<<endl;
        }

    }
    return 0;
}