HDU 1240.Asteroids!

Asteroids!

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

SubmitStatusPracticeHDU 1240

Description

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

Input

Input 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 present 

Starting 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. 

Output

For 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 1 O 0 0 0 0 0 0 END START 3 XXX XXX XXX OOO OOO OOO XXX XXX XXX 0 0 1 2 2 1 END START 5 OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO XXXXX XXXXX XXXXX XXXXX XXXXX OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO 0 0 0 4 4 4 END

Sample Output

1 0 3 4 NO ROUTE

 

 

三维空间中的BFS

 

难点不在于算法……在于读入（我觉得）

 

由于前面有START 后面有END

因此直接读入会产生问题

 

因此换用了自己写的函数来读入数据

 

/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o<n;o++)

const int maxn = 11;
int n;
int dis[maxn][maxn][maxn];
char Map[maxn][maxn][maxn];

const int delta[6][3] = {{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};

struct point {
    int x,y,z;
    point() {
        x = y = z = -1;
    }
    point(int a,int b,int c) {
        x = a;
        y = b;
        z = c;
    }
    bool operator == (const point &rhs)const {
        return ((x == rhs.x) && (y == rhs.y) && (z == rhs.z));
    }
};

inline int read_int() {
    char c;
    int ans = 0;
    while (c = getchar(),!(c >= '0' && c <= '9'));
    while (c >= '0'&&c <= '9') {
        ans *= 10;
        ans += (int)c - '0';
        c = getchar();
    }
    return ans;
}

int BFS(point s,point v) {
    if (s == v)
        return 0;

    memset(dis,-1,sizeof(dis));
    queue<point> Q;

    Q.push(s);
    dis[s.x][s.y][s.z] = 0;

    while (!Q.empty()) {
        int x = Q.front().x;
        int y = Q.front().y;
        int z = Q.front().z;
        Q.pop();

        REP(6) {
            int xx = x + delta[o][0];
            int yy = y + delta[o][1];
            int zz = z + delta[o][2];

            //非法路径
            if (xx < 0 || xx >= n || yy < 0 || yy >= n || zz < 0 || zz >= n)
                continue;
            //墙
            if (Map[xx][yy][zz] == 'X')
                continue;

            //尚未访问过
            if (dis[xx][yy][zz] == -1) {
                dis[xx][yy][zz] = dis[x][y][z] + 1;
                //到达终点
                if (point(xx,yy,zz) == v)
                    return dis[xx][yy][zz];
                Q.push(point(xx,yy,zz));
            }
        }
    }
    return -1;
}

bool Do() {
    char c;
    if (scanf("
%c",&c) == EOF)
        return false;
    n = read_int();
    //printf(" (%d) 
",n);

    for (int k = 0;k < n;k++)//块
        for (int i = 0;i < n;i++)//行
                scanf("%s",Map[k][i]);

    int s1,s2,s3,v1,v2,v3;
    s1 = read_int();
    s2 = read_int();
    s3 = read_int();
    v1 = read_int();
    v2 = read_int();
    v3 = read_int();
    //scanf("%d%d%d",&s1,&s2,&s3);
    //scanf("%d%d%d",&v1,&v2,&v3);
    point s = point(s3,s1,s2);
    point v = point(v3,v1,v2);


    

    int ans = BFS(s,v);

    if (ans == -1)
        printf("NO ROUTE
");
    else
        printf("%d %d
",n,ans);

    scanf("%*s");

    return true;
}

int main() {
    while (Do());
    return 0;
}