骑士问题--特殊的bfs

 这个是来自 国际大学生程序设计竞赛例题解（1）的题目

很简单的一道搜索题目，但是没有使用dfs，

使用的是特殊的bfs，非常巧妙

代码

/*
Qi Shi Wen Ti
 similar to eight queens
 use bfs() to get the min step
 */

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int queue[1000][2];

int matrix[8][8];
int startX,startY,endX,endY;
int head = -1,tail = 0;
int dir[8][2] = {{-1,-2},{-1,2},{-2,-1},{-2,1},{1,-2},{1,2},{2,-1},{2,1}};
void put(int x,int y)
{
	tail ++;
	if(tail >= 1000)
		tail = tail % 1000;
	queue[tail][0] = x;
	queue[tail][1] = y;
}
void get(int * x,int * y)
{
	head ++;
	if(tail < head)
	{
		printf("queue error!\n");
		exit(0);
	}	
	*x = queue[head][0];
	*y = queue[head][1];		
}
int queueEmpty()
{
	return tail < head ? 1:0; 
}
/* 
	the problem is how to calculate the step
	it seems like it's hard to find out one step's next step in the bfs search
	while it is easy in the dfs search;

    standard bfs
*/
void bfs(int startX,int startY,int endX,int endY)
{
	int curX,curY;
	int minStep=0,step=0;;
	put(startX,startY);
	while(! queueEmpty())
	{
		get(&curX,&curY);
		matrix[curX][curY] = 1;
		if(curX == endX && curY == endY)
		{
			if(step > minStep)
				minStep = step;
		}
		else if( 1) 
		{
			
		}
		else
		{
			for(int j=0;j<8;j++)
			{
				curX += dir[j][0];
				curY += dir[j][1];
				if(matrix[curX][curY] != 1 && curX >= 0 && curX < 8 && curY > -1 && curY < 8)
					put(curX,curY);
			}
		}
	}
}
/*
	special Bfs 
	it visited all nodes in a layer (just a layer's node int bfs tree) in a round
*/

int specialBfs(int startX,int startY,int endX,int endY)
{
	int head =0,tail =1,tmpTail;
	int step = 0;
	int curX = startX,curY = startY;

	queue[head][0] = startX;
	queue[head][1] = startY;
	matrix[curX][curY] = 1;

	while(head <tail)
	{
		step ++;
		tmpTail = tail;
		for(int i=head;i<tail;i++) //一次访问 整个树的一层节点，非常巧妙
		{
			for(int j=0;j<8;j++)
			{
				curX = queue[i][0] + dir[j][0];
				curY = queue[i][1] + dir[j][1];
				if(matrix[curX][curY] != 1 && curX >= 0 && curX < 8 && curY > -1 && curY < 8)
				{					
					queue[tmpTail][0] = curX;
					queue[tmpTail][1] = curY;
					tmpTail ++; // here we must add tmpTail after put node  in queue
					matrix[curX][curY] = 1;
					if(curX == endX && curY == endY)
						return step;
				}
			}
		}
		head = tail;
		tail = tmpTail; 
	}
	return -1;
}
int main()
{
	freopen("in.txt","r",stdin);
	int barrierCount;
	char c1,c2;
	scanf("%d\n",&barrierCount);
	for(int i=0;i<barrierCount;i++)
	{
		scanf("%c%c ",&c1,&c2);
		matrix[c1-'a'][c2-'1'] = 1;
	}
	scanf("%c%c\n",&c1,&c2);
	startX = c1 - 'a';
	startY = c2 - '1';
	scanf("%c%c\n",&c1,&c2);
	endX = c1 - 'a';
	endY = c2 - '1';
	printf("%d\n",specialBfs(startX,startY,endX,endY));
	fclose(stdin);
	return 0;
}