我只會土法煉鋼，時間複雜度直接O(m^2)

有人能給一下策略嗎

我原本的代碼:

package a;

import java.util.*;

public class i401 {

public static void main(String [] args) {

Scanner sc=new Scanner(System.in);

List<sign>map=new ArrayList<>();

map.add(new sign(0,0,0));

int n=sc.nextInt();

sc.nextLine();

for(int i=0;i<n;i++) {

String[]s=sc.nextLine().split(" ");

map.add(new sign(Integer.parseInt(s[0]),Integer.parseInt(s[1]),Integer.parseInt(s[2])));

}

int lastway=1;

sign nows=map.get(0);

int time=0;

while(nows!=null) {

time++;

if(nows.go==0) {

switch(lastway) {

case(1):lastway=2;break;

case(2):lastway=1;break;

case(3):lastway=4;break;

case(4):lastway=3;break;

}

}

else {

switch(lastway) {

case(1):lastway=4;break;

case(2):lastway=3;break;

case(3):lastway=2;break;

case(4):lastway=1;break;

}

}

sign temp=null;

int close=Integer.MAX_VALUE;

switch(lastway) {

case(1):

for(sign e:map) {

if(e.x==nows.x&&e.y>nows.y&&close>Math.abs(nows.y-e.y)) {

close=Math.abs(nows.y-e.y);

temp=e;

}

}

break;

case(2):

for(sign e:map) {

if(e.y==nows.y&&e.x>nows.x&&close>Math.abs(nows.x-e.x)) {

close=Math.abs(nows.x-e.x);

temp=e;

}

}

break;

case(3):

for(sign e:map) {

if(e.x==nows.x&&e.y<nows.y&&close>Math.abs(nows.y-e.y)) {

close=Math.abs(nows.y-e.y);

temp=e;

}

}

break;

case(4):

for(sign e:map) {

if(e.y==nows.y&&e.x<nows.x&&close>Math.abs(nows.x-e.x)) {

close=Math.abs(nows.x-e.x);

temp=e;

}

}

break;

}

nows=temp;

}

System.out.println(time);

sc.close();

}

}

class sign {

int x;

int y;

int go;

public sign(int x,int y,int go) {

this.x=x;

this.y=y;

this.go=go;

}

}