POJ 1915-Knight Moves（BFS-跳马）

小灰灰 2022-06-17 13:28 102阅读 0赞

Knight Moves

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;30000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;26034</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;12276</td> 
  </tr> 
 </tbody> 
</table>

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.  
![1915_1.jpg][]

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

## 题目意思： ##

给定棋盘的大小N和起终点两个位置的坐标，计算马从起点到终点所需的最小步骤。

## 解题思路： ##

跳马的问题做了不少，比如[POJ 2243][]，和这题特别像。

以前都是DFS水过去的，结果这个T了，┑(￣Д ￣)┍手写了个BFS才又水过去……

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <bitset>
    #include <queue>
    #include <iomanip>
    #include <algorithm>
    #define MAXN 333
    #define INF 0xfffffff
    using namespace std;
    bool vis[MAXN][MAXN];
    int step[MAXN][MAXN];
    int dir[8][2]= {
        {-2,-1},{-2,1},{-1,-2},{-1,2},{2,-1},{2,1},{1,-2},{1,2}};
    int res=0,st,ans;
    int sx,sy,ex,ey;
    int n;
    struct node
    {
        int x,y;
    };
    queue<node> q;
    void bfs(int x,int y)
    {
        node t;
        t.x=x,t.y=y;
        q.push(t);
        while(!q.empty())
        {
            t=q.front();
            q.pop();
            if(t.x==ex&&t.y==ey)
            {
                ans=step[t.x][t.y];
                return ;
            }
            for(int i=0; i<8; ++i)
            {
                node te;
                te.x=t.x+dir[i][0];
                te.y=t.y+dir[i][1];
                if(te.x<0||te.y<0||te.x>=n||te.y>=n||st>=ans) continue;
                {
                    if(!vis[te.x][te.y])
                    {
                        vis[te.x][te.y]=true;
                        step[te.x][te.y]=step[t.x][t.y]+1;
                        q.push(te);
                    }
                }
            }
        }
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("G:/cbx/out.txt","w",stdout);
    #endif
        /*ios::sync_with_stdio(false);
        cin.tie(0);*/
        int t;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
            memset(vis,false,sizeof(vis));
            memset(step,0,sizeof(step));
            while(!q.empty()) q.pop();
            st=0;
            ans=MAXN;
            bfs(sx,sy);
            printf("%d\n",ans);
        }
        return 0;
    }

[1915_1.jpg]: /images/20220617/39728a86c6ca4b109ce6670d277347ac.png
[TUD Programming Contest 2001]: http://poj.org/searchproblem?field=source&key=TUD+Programming+Contest+2001
[POJ 2243]: http://blog.csdn.net/mikasa3/article/details/52065240