POJ-1915 Knight Moves

Knight Moves

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 26185 Accepted: 12342

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

跳马题又双叒叕来了(ﾉ･ω･)ﾉﾞ，题意，给出起点和终点，马可以按照如图所示的八种方法跳跃，问马最少需要多少次才能跳到终点，如果无法到达则输出0。

思路：

还是BFS，创建一个step数组存储步数，开始初始化为0，若到达不了该点的step值为0，创建一个visit数组，用于存储该点是否被访问过，对于每个点，分别朝八个方向跳，如果坐标合法且没访问过，则加入队列，步数在原有基础上+1，修改此位置为已访问，否则不操作。最后输出结果即可。

#include<iostream>#include<cstring>#include<queue>#define MAX 400using namespace std;int dir[8][2]={{1,2},{1,-2},{2,1},{2,-1},{-1,2},{-1,-2},{-2,1},{-2,-1}};int sx,sy,ex,ey;int i;int n;int ans;bool visit[MAX][MAX];int step[MAX][MAX];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(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)    continue;            if(!visit[te.x][te.y])            {                visit[te.x][te.y]=true;                step[te.x][te.y]=step[t.x][t.y]+1;                q.push(te);            }        }    }}int main(){    int t;    cin >>t;    while(t--)    {        cin >>n;        cin >>sx>>sy>>ex>>ey;        while(!q.empty())   q.pop();        memset(step,0,sizeof(step));        memset(visit,false,sizeof(visit));        bfs(sx,sy);        cout << ans << endl;    }    return 0;}