HDU

2017.12.6对BFS的系列练习的第三题。
传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1240
Asteroids!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5587 Accepted Submission(s): 3550

Problem Description
You’re in space.
You want to get home.
There are asteroids.
You don’t want to hit them.

InputInput to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.A single data set has 5 components:Start line - A single line, "START N", where 1 <= N <= 10.Slice list - A series of N slices. Each slice is an N x N matrix representing a horizontal slice through the asteroid field. Each position in the matrix will be one of two values:'O' - (the letter "oh") Empty space'X' - (upper-case) Asteroid presentStarting Position - A single line, "A B C", denoting the <A,B,C> coordinates of your craft's starting position. The coordinate values will be integers separated by individual spaces.Target Position - A single line, "D E F", denoting the <D,E,F> coordinates of your target's position. The coordinate values will be integers separated by individual spaces.End line - A single line, "END"The origin of the coordinate system is <0,0,0>. Therefore, each component of each coordinate vector will be an integer between 0 and N-1, inclusive.The first coordinate in a set indicates the column. Left column = 0.The second coordinate in a set indicates the row. Top row = 0.The third coordinate in a set indicates the slice. First slice = 0.Both the Starting Position and the Target Position will be in empty space.OutputFor each data set, there will be exactly one output set, and there will be no blank lines separating output sets.A single output set consists of a single line. If a route exists, the line will be in the format "X Y", where X is the same as N from the corresponding input data set and Y is the least number of moves necessary to get your ship from the starting position to the target position. If there is no route from the starting position to the target position, the line will be "NO ROUTE" instead.A move can only be in one of the six basic directions: up, down, left, right, forward, back. Phrased more precisely, a move will either increment or decrement a single component of your current position vector by 1.

Sample Input

START 1O0 0 00 0 0ENDSTART 3XXXXXXXXXOOOOOOOOOXXXXXXXXX0 0 12 2 1ENDSTART 5OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOXXXXXXXXXXXXXXXXXXXXXXXXXOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO0 0 04 4 4END

Sample Output

1 03 4NO ROUTE

这种题意很坑啊，说有１００多组数据，又不说清楚，搞得我用循环１００次交ＷＡ两次，后面果断EOF。。。
也是挺轻松的三维BFS，注意一下ｘ的方向选取。
AC代码如下：

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int maxn = 15;char space[maxn][maxn][maxn];int vis[maxn][maxn][maxn];struct Node{    int x, y, z;    int step;};int n;int a, b, c;int d, e, f;int dx[6] = {1, -1, 0, 0, 0, 0};int dy[6] = {0, 0, 1, -1, 0, 0};int dz[6] = {0, 0, 0, 0, 1, -1};int BFS(void){    memset(vis, 0, sizeof(vis));    queue<Node> craft;    Node beg;    beg.x = c, beg.y = a, beg.z = b, beg.step = 0;    craft.push(beg);    vis[beg.x][beg.y][beg.z] = 1;    while(!craft.empty())    {        Node cur = craft.front();        craft.pop();        if(cur.x == f && cur.y == d && cur.z == e)    return cur.step;        for(int i = 0; i < 6; i++)        {            Node next;            next.x = cur.x + dx[i];            next.y = cur.y + dy[i];            next.z = cur.z + dz[i];            next.step = cur.step + 1;            if(next.x < 0 || next.x >= n || next.y < 0 || next.y >= n || next.z < 0 || next.z >= n)    continue;            if(vis[next.x][next.y][next.z] || space[next.x][next.y][next.z] == 'X')    continue;            if(next.x == f && next.y == d && next.z == e)                return next.step;            vis[next.x][next.y][next.z] = 1;            craft.push(next);        }    }    return -1;}int main(){    char s[maxn];    while(scanf("%s %d", s, &n) != EOF && n)    {        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++)                scanf("%s", space[i][j]);        scanf("%d %d %d", &a, &b, &c);        scanf("%d %d %d", &d, &e, &f);        scanf("%s", s);        int flag = BFS();        if(flag == -1)            printf("NO ROUTE\n");        else            printf("%d %d\n", n, flag);    }    return 0;}