poj2243 && hdu1372 Knight Moves（BFS）

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题目链接：

POJ：http://poj.org/problem?id=2243

HDU: http://acm.hdu.edu.cn/showproblem.php?

pid=1372

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.

题意：用象棋中跳马的走法。从起点到目标点的最小步数；

代码例如以下：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
#define M 1017
struct node
{
	int x, y;
	int step;
};
int xx[8] = {1,2,2,1,-1,-2,-2,-1};
int yy[8] = {2,1,-1,-2,-2,-1,1,2};
bool vis[M][M];
int n, ansx, ansy;
queue<node>q;
int BFS(int x, int y)
{
	
	if(x == ansx && y == ansy)
		return 0;
	int dx, dy, i;
	node front, rear;
	front.x = x, front.y = y, front.step = 0;
	q.push(front);
	vis[x][y] = true;
	while(!q.empty())
	{
		front = q.front();
		q.pop();
		for(i = 0; i < 8; i++)
		{
			dx = front.x+xx[i];
			dy = front.y+yy[i];
			if(dx>=1&&dx<=8&&dy>=1&&dy<=8&&!vis[dx][dy])
			{
				vis[dx][dy] = true;
				if(dx == ansx && dy == ansy)
				{
					return front.step+1;
				}
				rear.x = dx, rear.y = dy, rear.step = front.step+1;
				q.push(rear);
			}
		}
	}

}
int main()
{
	char s,e;
	int a1,a2;
	while(~scanf("%c%d %c%d",&s,&a1,&e,&a2))
	{
		getchar();
		while(!q.empty())
			q.pop();
		memset(vis,0,sizeof(vis));
		int s1 = s-'a'+1;
		int e1 = e-'a'+1;
		ansx = e1, ansy = a2;
		int ans = BFS(s1,a1);
		printf("To get from %c%d to %c%d takes %d knight moves.\n",s,a1,e,a2,ans);
	}
	return 0;
}