HDU 1240 (简单三维广搜) Asteroids!

给出一个三维的迷宫以及起点和终点，求能否到大终点，若果能输出最短步数

三维的问题无非就是变成了6个搜索方向

最后强调一下xyz的顺序，从输入数据来看，读入的顺序是map[z][x][y]

总之，这是很基础的一道题

//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

struct Point
{
    char type;
    int x, y, z;
    int steps;
}start, end;

char map[12][12][12];
int dir[6][3] = {{1,0,0}, {-1,0,0}, {0,1,0}, {0,-1,0}, {0,0,1}, {0,0,-1}};
char buff[10];
int n;

bool islegal(int x, int y, int z)
{
    return (x>=0 && x<n && y>=0 && y<n && z>=0 && z<n && map[z][x][y]!='X');
}

void BFS(void)
{
    queue<Point> qu;
    start.steps = 0;
    qu.push(start);
    while(!qu.empty())
    {
        Point now = qu.front();
        if(now.x==end.x && now.y==end.y && now.z==end.z)
        {
            printf("%d %d
", n, now.steps);
            return;
        }
        for(int i = 0; i < 6; ++i)
        {
            int xx = now.x + dir[i][0];
            int yy = now.y + dir[i][1];
            int zz = now.z + dir[i][2];
            if(islegal(xx, yy, zz))
            {
                Point next;
                next.x = xx, next.y = yy, next.z = zz, next.steps = now.steps + 1;
                map[zz][xx][yy] = 'X';
                qu.push(next);
            }
        }
        qu.pop();
    }
    printf("NO ROUTE
");
}

int main(void)
{
    #ifdef LOCAL
        freopen("1240in.txt", "r", stdin);
    #endif

    while(cin >> buff >> n)
    {
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)
                cin >> map[i][j];

        cin >> start.x >> start.y >> start.z;
        cin >> end.x >> end.y >> end.z;
        cin >> buff;
        BFS();
    }
    return 0;
}