POJ 1915 Knight Moves (BFS)

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本文介绍了一种计算棋盘上骑士从起点到终点最短步数的算法。通过使用广度优先搜索(BFS)，该算法能够找出骑士到达目标位置所需的最少移动次数。文章提供了完整的C++代码实现，包括输入输出示例。

Problem 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

The input begins with the number n of scenarios on a single line by itself.
Next follow n scenarios. Each scenario consists of three lines containing integer numbers. The first line specifies the length l of a side of the chess board (4 <= l <= 300). The entire board has size l * l. The second and third line contain pair of integers {0, ..., l-1}*{0, ..., l-1} specifying the starting and ending position of the knight on the board. The integers are separated by a single blank. You can assume that the positions are valid positions on the chess board of that scenario.

Output

For each scenario of the input you have to calculate the minimal amount of knight moves which are necessary to move from the starting point to the ending point. If starting point and ending point are equal,distance is zero. The distance must be written on a single line.

Sample Input

Sample Output

5
28
0

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
 
int n,sx,sy,ex,ey;
int vis[305][305];
int to[8][2] = {-1,-2,-2,-1,-2,1,-1,2,1,-2,2,-1,2,1,1,2};
struct node
{
    int x,y,step;
};
 
int check(int x,int y)
{
    if(x<0 || x>=n || y<0 || y>=n)
    return 1;
    return 0;
}
 
void bfs()
{
    memset(vis,0,sizeof(vis));
    node a,next;
    queue<node> Q;
    int i;
    a.x = sx;
    a.y = sy;
    a.step = 0;
    vis[sx][sy] = 1;
    Q.push(a);
    while(!Q.empty())
    {
        a = Q.front();
        Q.pop();
        if(a.x == ex && a.y == ey)
        {
            printf("%d\n",a.step);
            return;
        }
        for(i = 0;i<8;i++)
        {
            next = a;
            next.x = a.x+to[i][0];
            next.y = a.y+to[i][1];
            if(check(next.x,next.y))
            continue;
            next.step = a.step+1;
            vis[next.x][next.y] = 1;
            Q.push(next);
        }
    }
}
 
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d%d",&n,&sx,&sy,&ex,&ey);
        bfs();
    }
 
    return 0;
}