Problem:
Analysis:
This problem is a depth-first-search (DFS) algorithm, however, we can not implement it recursively. Because it allows 10000 steps, which lets us save too many states.
We can implement it iteratively because at each location, there is only one choice for the robot to move according to the rule, then we don't need to save the previous states,
There are four direction, at each location, we decide its face direction first, and then start from its left hand direction to find whether there is a way.
Time complexity is O(1), since no more than 10000 steps.
My solution: (Your opinion is highly appreciated)
package codeJam.google.com;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
/**
* @author Zhenyi 2013 Dec 21, 2013 12:47:00 PM
*/
enum DIRECTION {
E, S, W, N
};
public class CrossTheMaze {
static String path = "";
static Integer ex;
static Integer ey;
static Integer sx;
static Integer sy;
static Integer steps;
static char[][] maze;
static int N;
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new FileReader(
"C:/Users/Zhenyi/Downloads/D-small-practice.in"));
FileWriter out = new FileWriter(
"C:/Users/Zhenyi/Downloads/D-small-practice.out");
// BufferedReader in = new BufferedReader(new
// FileReader("C:/Users/Zhenyi/Downloads/D-large-practice.in"));
// FileWriter out = new
// FileWriter("C:/Users/Zhenyi/Downloads/D-large-practice.out");
int T = new Integer(in.readLine());
for (int cases = 1; cases <= T; cases++) {
N = new Integer(in.readLine());
maze = new char[N][N];
for (int i = 0; i < N; i++) {
maze[i] = in.readLine().toCharArray();
}
String[] st = in.readLine().split("\\s");
sx = new Integer(st[0]) - 1;
sy = new Integer(st[1]) - 1;
ex = new Integer(st[2]) - 1;
ey = new Integer(st[3]) - 1;
Integer faceDirc = 0;
if (sx == N - 1 && sy == 0) {
faceDirc = 1;
}
if (sx == N - 1 && sy == N - 1) {
faceDirc = 2;
}
if (sx == 0 && sy == N - 1) {
faceDirc = 3;
}
path = "";
steps = 0;
if (findPath(faceDirc)) {
out.write("Case #" + cases + ": " + steps + "\n");
out.write(path + "\n");
} else {
out.write("Case #" + cases + ": " + "Edison ran out of energy."
+ "\n");
}
}
in.close();
out.flush();
out.close();
}
private static boolean findPath(Integer faceDirc) {
boolean found = false;
while (steps <= 10000 && !found) {
if (sx == ex && sy == ey) {
found = true;
break;
}
int startDirc = (faceDirc + 1 <= 3) ? (faceDirc + 1)
: (faceDirc - 3);
boolean findway = false;
for (int i = 0; i < 4 && !findway; i++) {
int num = (startDirc - i) % 4 >= 0 ? (startDirc - i)
: (startDirc - i + 4);
switch (num) {
case 0:
if (sx + 1 < N && maze[sx + 1][sy] == '.') {
findway = true;
path = path + "S";
steps++;
faceDirc = 0;
sx++;
}
break;
case 1:
if (sy + 1 < N && maze[sx][sy + 1] == '.') {
findway = true;
path = path + "E";
steps++;
faceDirc = 1;
sy++;
}
break;
case 2:
if (sx - 1 >= 0 && maze[sx - 1][sy] == '.') {
findway = true;
path = path + "N";
steps++;
faceDirc = 2;
sx--;
}
break;
case 3:
if (sy - 1 >= 0 && maze[sx][sy - 1] == '.') {
findway = true;
path = path + "W";
steps++;
faceDirc = 3;
sy--;
}
break;
}
}
if (!findway) {
steps = 20000;
}
}
// TODO Auto-generated method stub
return found;
}
}
作者：u013301594 发表于2013-12-28 0:18:08 原文链接
阅读：193 评论：0 查看评论