(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
原文地址:https://www.cnblogs.com/whywhy/p/4228073.html