Knight Moves(POJ1915)

 

 

 

 通过广搜, 向如图所示的8个方向搜索, 边搜索边记录步数, 最后若到达终点则返回当前走过的步数, 否则返回0

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;

struct node
{
	int x, y;	// 当前点的坐标 
	int step;	// 当前走的步数
	node() {}
	node(int x1, int y1):x(x1), y(y1) {} 
};

int n, l;
int map[310][310], vis[310][310];
int dir[8][2] = {{2,1},{1,2},{-1,2},{-2,1},{-1,-2},{-2,-1},{1,-2},{2,-1}};
node s, e;

int bfs()
{
	queue<node> q;
	node p, t;
	int dx, dy;
	s.step = 0;
	vis[s.x][s.y] = 1;
	q.push(s);
	while(!q.empty())
	{
		p = q.front();
		q.pop();
		
		if(p.x == e.x && p.y == e.y)	return p.step;
		
		for(int i = 0; i < 8; ++ i)
		{
			dx = p.x + dir[i][0];
			dy = p.y + dir[i][1];
			
			if(dx < 0 || dx >= l || dy < 0 || dy >= l || vis[dx][dy])	
				continue;
				
			t = node(dx, dy);
			t.step = p.step + 1;
			vis[t.x][t.y] = 1;
			q.push(t);
		}
	}
	return 0;
}

int main()
{
	cin >> n;
	while(n --)
	{
		memset(vis, 0, sizeof(vis));
		cin >> l;
		cin >> s.x >> s.y;
		cin >> e.x >> e.y;
		cout << bfs() << endl;
	}
	
	return 0;
}

/*
Sample Input:
3
8
0 0
7 0
100
0 0
30 50
10
1 1
1 1

Sample Output:
5
28
0
*/