HDU2612 Find a way （雙廣搜）

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki. 
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. 
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes. 

InputThe input contains multiple test cases. 
Each test case include, first two integers n, m. (2<=n,m<=200). 
Next n lines, each line included m character. 
‘Y’ express yifenfei initial position. 
‘M’    express Merceki initial position. 
‘#’ forbid road; 
‘.’ Road. 
‘@’ KCF 
OutputFor each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.Sample Input

4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

Sample Output

66
88
66

題意：求2個點到同一個點的最短路程

注意：某些點可能兩個點不能同時到達，因此要用兩個數組分別記錄兩次廣搜的路徑到達狀況

仍是強調多個輸入，每次都要數組初始化、隊列清空

代碼：

import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Scanner;
class Node{
        int x;
        int y;
        int step;
        public Node(int x,int y,int step){
                this.x=x; this.y=y; this.step=step;
        }
}
public class Main{
        static final int N=205;
        static int n,m;
        static int dx[]={0,0,1,-1};
        static int dy[]={1,-1,0,0};
        static char map[][]=new char[N][N];
        static int s1[][]=new int[N][N];
        static int s2[][]=new int[N][N];
        static boolean vis[][]=new boolean[N][N];
        static ArrayDeque<Node> q=new ArrayDeque<Node>();
        static void init(){
                  for(int i=0;i<N;i++)  Arrays.fill(s1[i], 0);
                  for(int i=0;i<N;i++)  Arrays.fill(s2[i], 0);
        }
        static void bfs(int sx,int sy,int s[][]){
                while(!q.isEmpty()) q.poll();
                vis[sx][sy]=true;
                q.offer(new Node(sx,sy,0));
                while(!q.isEmpty()){
                        Node t=q.poll();
                        for(int i=0;i<4;i++){
                                int xx=t.x+dx[i];
                                int yy=t.y+dy[i];
                                if(xx<0||yy<0||xx>=n||yy>=m ||vis[xx][yy]||map[xx][yy]=='#') continue;
                                vis[xx][yy]=true;
                                s[xx][yy]=t.step+1;
                                q.offer(new Node(xx,yy,t.step+1));
                        }
                        
                }
        }
         public static void main(String[] args) {
                 Scanner scan=new Scanner(System.in);
                 while(scan.hasNext()){
                         n=scan.nextInt();
                         m=scan.nextInt();
                         for(int i=0;i<n;i++) map[i]=scan.next().toCharArray();

                         int yx=0,yy=0,mx=0,my=0;
                         for(int i=0;i<n;i++)
                             for(int j=0;j<m;j++)
                                 if(map[i][j]=='Y'){
                                         yx=i; yy=j;
                                 }
                                 else if(map[i][j]=='M'){
                                         mx=i; my=j;
                                 }
                         
                         init();
                         for(int i=0;i<N;i++)  Arrays.fill(vis[i], false);
                         bfs(yx,yy,s1);
                         for(int i=0;i<N;i++)   Arrays.fill(vis[i], false);
                         bfs(mx,my,s2);
                         
                         int min=2147483647;
                         for(int i=0;i<n;i++)
                             for(int j=0;j<m;j++)
                                 if(map[i][j]=='@' && s1[i][j]!=0 &&s2[i][j]!=0)
                                       min=Math.min(min, s1[i][j]+s2[i][j]);
                         System.out.println(min*11);
                 }
        }
 }