• POJ 2251 Dungeon Master 三维bfs


    题目:http://poj.org/problem?id=2251

    题意:输入一个L层的R*C迷宫,从S点出发,判断能否到达E,如果能,输出最短时间。

    解法:简单bfs,把四个方向向量换成六个即可。

    AC:

    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <queue>
    #include <algorithm>
    using namespace std;
    
    char map[35][35][35];
    int vis[35][35][35];
    int k,n,m,sx,sy,sz,ex,ey,ez;
    int to[6][3] = {{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}};
    
    struct node
    {
        int x,y,z,step;
    };
    
    int check(int x,int y,int z)
    {
        if(x<0 || y<0 || z<0 || x>=k || y>=n || z>=m)
            return 1;
        else if(map[x][y][z] == '#')
            return 1;
        else if(vis[x][y][z])
            return 1;
        return 0;
    }
    
    int bfs()
    {
        int i;
        node a,next;
        queue<node> Q;
        a.x = sx,a.y = sy,a.z = sz;
        a.step = 0;
        vis[sx][sy][sz] = 1;
        Q.push(a);
        while(!Q.empty())
        {
            a = Q.front();
            Q.pop();
            if(a.x == ex && a.y == ey && a.z == ez)
                return a.step;
            for(i = 0; i<6; i++)
            {
                next = a;
                next.x = a.x+to[i][0];
                next.y = a.y+to[i][1];
                next.z = a.z+to[i][2];
                if(check(next.x,next.y,next.z))
                    continue;
                vis[next.x][next.y][next.z] = 1;
                next.step = a.step+1;
                Q.push(next);
            }
        }
        return 0;
    }
    
    int main()
    {
        int i,j,r;
        while(scanf("%d%d%d",&k,&n,&m),n+m+k)
        {
            for(i = 0; i<k; i++)
            {
                for(j = 0; j<n; j++)
                {
                    scanf("%s",map[i][j]);
                    for(r = 0; r<m; r++)
                    {
                        if(map[i][j][r] == 'S')
                        {
                            sx = i,sy = j,sz = r;
                        }
                        else if(map[i][j][r] == 'E')
                        {
                            ex = i,ey = j,ez = r;
                        }
                    }
                }
            }
            memset(vis,0,sizeof(vis));
            int ans;
            ans = bfs();
            if(ans)
                printf("Escaped in %d minute(s).
    ",ans);
            else
                printf("Trapped!
    ");
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zz990728/p/8900512.html
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