poj1915 Knight Move

Knight Moves

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 22762 Accepted: 10636

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

Source

TUD Programming Contest 2001, Darmstadt, Germany

题解：

给定起始和终止位置，求最小步数，显然BFS。为了节省时间要选择双向BFS。每次选择节点少的开始扩展。

程序代码：

<span style="font-size:18px;">#include<iostream>#include<cstdio>#include<cstring>#define MAXN 100000using namespace std;struct node{int x,y;}q[2][MAXN];int num,ans,n,l[2],r[2],dis[2][305][305];bool v[2][305][305];int epdx[8]={1,2,2,1,-1,-2,-2,-1},epdy[8]={2,1,-1,-2,-2,-1,1,2};int expand(int k){int i,j,x,y,d,tmpx,tmpy;x=q[k][l[k]].x;y=q[k][l[k]].y;d=dis[k][x][y];for (i=0; i<8; i++){tmpx=x+epdx[i];tmpy=y+epdy[i];if (tmpx>=0&&tmpx<n&&tmpy>=0&&tmpy<n&&!v[k][tmpx][tmpy]){v[k][tmpx][tmpy]=true;r[k]++;q[k][r[k]].x=tmpx;q[k][r[k]].y=tmpy;dis[k][tmpx][tmpy]=d+1;if (v[1-k][tmpx][tmpy]){ans=dis[k][tmpx][tmpy]+dis[1-k][tmpx][tmpy];return 1;}}}return 0;}void bfs(){if (q[0][1].x==q[1][1].x&&q[0][1].y==q[1][1].y) return;v[0][q[0][1].x][q[0][1].y]=true;v[1][q[1][1].x][q[1][1].y]=true;l[0]=r[0]=l[1]=r[1]=1;while (l[0]<=r[0]&&l[1]<=r[1]){if (r[0]-l[0]<r[1]-l[1]) {if(expand(0)) return; l[0]++;}else {if(expand(1)) return; l[1]++;}}}int main(){cin>>num;for (int i=1; i<=num; i++){memset(v,false,sizeof(v));memset(q,0,sizeof(q));memset(dis,0,sizeof(dis));ans=0;cin>>n;cin>>q[0][1].x>>q[0][1].y;cin>>q[1][1].x>>q[1][1].y;bfs();cout<<ans<<endl;}return 0;}//Written by aap</span>