[POJ1915] Knight Moves

最新推荐文章于 2022-12-21 20:33:47 发布

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本文介绍了一种计算国际象棋中骑士从一个位置移动到另一个位置所需的最少步数的算法。通过广度优先搜索（BFS）实现，适用于4到300大小的棋盘。输入包括棋盘尺寸及起点和终点坐标。

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.
这里写图片描述

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

Sample Output
5
28
0

Source
TUD Programming Contest 2001, Darmstadt, Germany

广搜，就是一个地图来回跳

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int t,n,sx,sy,ax,ay;
bool vis[4005][4005];
struct point{
    int x,y,step;
}node,r;
bool flag;
point pre[4005][4005];
int dx[8]={1,2,2,1,-1,-2,-2,-1},
    dy[8]={2,1,-1,-2,-2,-1,1,2};
void bfs(int x,int y)
{
    if(x==ax && y==ay)
    {
        cout<<"0"<<endl;
        return;
    }
    vis[x][y]=1;
    node.x=0;
    node.y=0;
    node.step=0;
    pre[x][y]=node;
    node.x=x;
    node.y=y;
    queue<point> Q;
    Q.push(node);
    while(!Q.empty() && !flag)
    {
        r=Q.front();
        Q.pop();
        int nx,ny;
        for(int i=0;i<=7;i++)
        {
            nx=r.x+dx[i];
            ny=r.y+dy[i];
            if(nx<0||nx>=n||ny<0||ny>=n)
                continue;
            if(vis[nx][ny])
                continue;
            vis[nx][ny]=1;
            pre[nx][ny]=r;
            node.x=nx;
            node.y=ny;
            node.step=r.step+1;
            Q.push(node);
            if(nx==ax&&ny==ay)
            {
                cout<<node.step<<endl;
                flag=1;
            }
        }
    }
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        flag=0;
        scanf("%d",&n);
        scanf("%d %d",&sx,&sy);
        scanf("%d %d",&ax,&ay);
        bfs(sx,sy);
    }
    return 0;
}