POJ 1915 Knight Moves

标签：acm   bfs   c++   

Description

Background 
Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him? 
The Problem 
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov. 
For people not familiar with chess, the possible knight moves are shown in Figure 1. 

Input

Output

Sample Input

3
8
0 0
7 0
100
0 0
30 50
10
1 1
1 1

Sample Output

5
28
0

代码：

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
struct point{
	int x;
	int y;
	int step;
}p, pp;
int d[8][2] = { { -2, 1 }, { -1, 2 }, { 1, 2 }, { 2, 1 }, { -2, -1 }, { -1, -2 }, { 1, -2 }, { 2, -1 } };
int x_1, y_1, x_2, y_2, l;
int vis[500][500];
queue<point>s;
bool isok(int x, int y)
{
	if (x >= 0 && x < l&&y >= 0 && y < l)
		return true;
	return false;
}
int bfs()
{
	if (x_1 == x_2&&y_1 == y_2)
		return 0;
	while (!s.empty())
		s.pop();
	p.x = x_1, p.y = y_1;
	p.step = 0;
	s.push(p);
	vis[x_1][y_1] = 1;
	while (!s.empty())
	{
		p = s.front();
		s.pop();
		if (p.x == x_2&&p.y == y_2)
			return p.step;
		for (int i = 0; i < 8; i++){
			int xx = p.x + d[i][0];
			int yy = p.y + d[i][1];
			if (isok(xx, yy) && !vis[xx][yy]){
				vis[xx][yy] = 1;
				pp.x = xx, pp.y = yy;
				pp.step = p.step + 1;
				s.push(pp);
			}
		}
	}
}
int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &l);
		scanf("%d%d", &x_1, &y_1);
		scanf("%d%d", &x_2, &y_2);
		memset(vis, 0, sizeof(vis));
		printf("%d\n", bfs());
	}
	return 0;
}

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POJ 1915 Knight Moves

标签：acm   bfs   c++   

原文地址：http://blog.csdn.net/u012964281/article/details/38144731