zoj 1438 - Asteroids!

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1438

还是很经典的搜索题，有意思的是把原来的二维迷宫改为了立体迷宫，用三维数组来存储。易错的是三维容易把坐标弄错，我因为这wa了好多次。。。在纸上多画画就ok。

代码如下：

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

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

int N;
char path[15][15][15];
char s[6];
int vis[15][15][15];
typedef struct
{
    int x;
    int y;
    int z;
    int step;
}State;
State begin, end;
queue<State> q;

int final(State s1, State s2)
{
    if(s1.x == s2.x && s1.y == s2.y && s1.z == s2.z)
        return 1;
    return 0;
}

int isright(State u)
{
    if(u.x >= 0 && u.x < N && u.y >= 0 && u.y < N && u.z >= 0 && u.z < N)
        return 1;
    return 0;
}

int main()
{
    int i, j, k, ok;
    while(scanf("%s%d", s, &N) != EOF)
    {
        getchar();
        for(i = 0; i < N; i++)
            for(j = 0; j < N; j++)
            {
                scanf("%s", path[i][j]);
            }
        scanf("%d%d%d", &begin.x, &begin.y, &begin.z);
        scanf("%d%d%d", &end.x, &end.y, &end.z);
        scanf("%s", s);
        memset(vis, 0, sizeof(vis));
        ok = 0;
        begin.step = 0;
        vis[begin.z][begin.y][begin.x] = 1;
        q.push(begin);
        while(!q.empty())
        {
            State u = q.front();
            q.pop();
            if(final(u, end))
            {
                ok = 1;
                printf("%d %d\n", N, u.step);
                break;
            }
            for(i = 0; i < 6; i++)
            {
                State v;
                v.step = u.step + 1;
                v.x = u.x + dx[i];
                v.y = u.y + dy[i];
                v.z = u.z + dz[i];
                if(path[v.z][v.y][v.x] != 'X' && isright(v) && !vis[v.z][v.y][v.x])
                {
                    q.push(v);
                    vis[v.z][v.y][v.x] = 1;
                }
            }
        }
        if(!ok)
            printf("NO ROUTE\n");
        while(!q.empty())
            q.pop();
    }
    return 0;
}