zoukankan      html  css  js  c++  java
  • (简单) POJ 2251 Dungeon Master,BFS。

      Description

      You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.

      Is an escape possible? If yes, how long will it take?

      一个三维的迷宫问题,和二维没什么区别,直接BFS就好。

    代码如下:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<queue>
    
    using namespace std;
    
    bool map1[40][40][40];
    bool vis[40][40][40];
    int L,R,C;
    int Sl,Sr,Sc,El,Er,Ec;
    
    struct state
    {
        int l,r,c;
        int num;
    
        state() {}
        state(int x,int y,int z,int n):l(x),r(y),c(z),num(n) {}
    };
    
    bool judge(int x,int y,int z)
    {
        if(x<=0||x>L||y<=0||y>R||z<=0||z>C)
            return 0;
    
        if(map1[x][y][z]==0)
            return 0;
    
        if(vis[x][y][z])
            return 0;
    
        vis[x][y][z]=1;
        return 1;
    }
    
    int bfs()
    {
        queue <state> que;
        state temp;
        int tl,tr,tc;
    
        que.push(state(Sl,Sr,Sc,0));
    
        while(!que.empty())
        {
            temp=que.front();
            que.pop();
    
            if(temp.l==El&&temp.r==Er&&temp.c==Ec)
                return temp.num;
    
            tl=temp.l;
            tr=temp.r;
            tc=temp.c;
    
            if(judge(tl-1,tr,tc))
                que.push(state(tl-1,tr,tc,temp.num+1));
            if(judge(tl+1,tr,tc))
                que.push(state(tl+1,tr,tc,temp.num+1));
            if(judge(tl,tr-1,tc))
                que.push(state(tl,tr-1,tc,temp.num+1));
            if(judge(tl,tr+1,tc))
                que.push(state(tl,tr+1,tc,temp.num+1));
            if(judge(tl,tr,tc-1))
                que.push(state(tl,tr,tc-1,temp.num+1));    
            if(judge(tl,tr,tc+1))
                que.push(state(tl,tr,tc+1,temp.num+1));
        }
    
        return -1;
    }
    
    int main()
    {
        char s[50];
        int ans;
    
        ios::sync_with_stdio(false);
    
        while(cin>>L>>R>>C)
        {
            memset(vis,0,sizeof(vis));
    
            if(!L&&!R&&!C)
                break;
    
            for(int i=1;i<=L;++i)
                for(int j=1;j<=R;++j)
                {
                    cin>>s;
                    for(int k=1;k<=C;++k)
                        switch(s[k-1])
                        {
                            case 'S':
                                Sl=i;
                                Sr=j;
                                Sc=k;
                                map1[i][j][k]=1;
                                break;
                            case 'E':
                                El=i;
                                Er=j;
                                Ec=k;
                                map1[i][j][k]=1;
                                break;
                            case '.':
                                map1[i][j][k]=1;
                                break;
                            case '#':
                                map1[i][j][k]=0;
                                break;
                        }
                }
    
            ans=bfs();
    
            if(ans!=-1)
                cout<<"Escaped in "<<ans<<" minute(s).
    ";
            else
                cout<<"Trapped!
    ";
        }
    
        return 0;
    }
    View Code
  • 相关阅读:
    0401-服务注册与发现、Eureka简介
    001-OSI七层模型,TCP/IP五层模型
    云原生应用开发12-Factors
    0301-服务提供者与服务消费者
    0201-开始使用Spring Cloud实战微服务准备工作
    0107-将Monolith重构为微服务
    0106-选择微服务部署策略
    0105-微服务的事件驱动的数据管理
    0104-微服务体系结构中的服务发现
    0103-微服务架构中的进程间通信
  • 原文地址:https://www.cnblogs.com/whywhy/p/4228073.html
Copyright © 2011-2022 走看看