POJ1915

Knight Moves

Time Limit: 1000MS
Memory Limit: 30000KTotal Submissions: 24777
Accepted: 11709

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

380 07 01000 030 50101 11 1

Sample Output

5280

bfs题，给出棋盘边长和骑士的起始坐标和终点坐标，求到达步数。

只是方向有点多，但基本做法还是和普通的bfs一致。

#include<iostream>#include<cstdio>#include<queue>#include<string.h>using namespace std;#define Maxn 305int vis[Maxn][Maxn];int dir[8][2]={{2,1},{1,2},{-1,2},{-2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}};int m;struct Node{    int x,y,d;    void init(int nx,int ny,int nd)    {        x=nx;        y=ny;        d=nd;    }};Node head,tail;int bfs(Node source , Node target){queue<Node> Q;source.d=0;Q.push(source);memset(vis,0,sizeof(vis));vis[source.x][source.y]=1;while(!Q.empty())        {Node a=Q.front();Q.pop();int x=a.x;int y=a.y;for(int i=0;i<8;i++)            {int nx = x + dir[i][0] ;int ny = y + dir[i][1] ;if(nx < 0 || nx >= m || ny >= m || ny < 0)continue ;if(vis[nx][ny])continue ;if(target.x == nx && target.y == ny)return a.d + 1 ;           vis[nx][ny] = 1 ;Node c ;c.init(nx , ny , a.d + 1) ;Q.push(c) ;}}return -1 ;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&m);        scanf("%d%d",&head.x,&head.y);        scanf("%d%d",&tail.x,&tail.y);        int tmp=bfs(head,tail);        if(tmp==-1) printf("0\n");        else printf("%d\n",tmp);    }    return 0;}