SRM 253 div2 1000pts

这题不难，用bfs，bfs的时候注意剪一个枝就行了。代码如下：

import java.io.*;
import java.util.*;
import java.util.regex.*;

//photo,print 
public class ABCPath {
   public int length(String[] grid) {
		int ans=0;
		int row=grid.length;
		int col=grid[0].length();
		char[][] chs=new char[row][];
		int[][] bchs=new int[row][col];
		LinkedList<Pos> ll=new LinkedList<Pos>();
		for(int i=0;i<row;i++){
			chs[i]=grid[i].toCharArray();
			Arrays.fill(bchs[i],0);
			for(int j=0;j<col;j++)
				if(chs[i][j]=='A'){
					ll.add(new Pos(i,j,'A'));
					bchs[i][j]=1;
				}
		}
		
		int step[][]=new int[][]{{0,1},{0,-1},{-1,0},{1,0},{-1,1},{-1,-1},{1,1},{1,-1}};
		while(ll.size()>0){
			Pos pos=ll.remove();
			for(int i=0;i<step.length;i++){
				int nrow=step[i][0]+pos.row;
				int ncol=step[i][1]+pos.col;
				if(nrow>=0&&nrow<row&&ncol>=0&&ncol<col){
					if(chs[nrow][ncol]==(char)(pos.c+1)){
						if(pos.length+1>bchs[nrow][ncol]){
							ll.add(new Pos(nrow,ncol,chs[nrow][ncol],pos.length+1));
							bchs[nrow][ncol]=pos.length+1;
						}
					}//if
				}//if
			}//for
		}//while
		
		for(int i=0;i<row;i++){
			for(int j=0;j<col;j++){
				if(bchs[i][j]>ans)
					ans=bchs[i][j];
			}
		}
		
		return ans;
   }
   
   class Pos{
   	int row;
   	int col;
   	char c;
   	int length=1;	
   	public Pos(int row,int col){
   		this.row=row;
   		this.col=col;
   	}
   	
   	public Pos(int row,int col,char c){
   		this(row,col);
   		this.c=c;
   	}
   	
   	public Pos(int row,int col,char c,int length){
   		this(row,col,c);
   		this.length=length;	
   	}
   }
}