Bfs1.4(两点之间最短距离)

按照象棋中马走的方向,判断两点间最短距离或不可到达;

http://poj.org/problem?id=2243

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int maxn=1001+5;
int dx[]={-2,-2,-1,-1,1,1,2,2};
int dy[]={-1,1,2,-2,2,-2,1,-1};
int sr,sc,er,ec;
int n,m;
struct Node
{
    int r,c,time;
    Node() {time=0;}
    Node(int r,int c,int k):r(r),c(c),time(k) {}
};
Node no;
int visit[100][100];
string a,b;
void  BFS(Node sta)
{
    visit[sta.r][sta.c]=1;
    queue<Node> Q;
    Q.push(Node(sta.r,sta.c,sta.time));
    while(!Q.empty())
    {
        Node x=Q.front();
        Q.pop();
        for(int i=0; i<8; i++)
        {
            int nr=x.r+dx[i];
            int nc=x.c+dy[i];
            if(nr>=0 && nr<=7 && nc>=0 && nc<=7 && !visit[nr][nc])
            {
                int time=x.time+1;
                if(nr==no.r&&nc==no.c){
                    cout<<"To get from "<<a<<" to "<<b<<" takes "<<time<<" knight moves."<<endl;
                    return ;
                }
                visit[nr][nc]=1;
                Q.push(Node(nr,nc,time));
            }
        }
    }
    cout<<"To get from "<<a<<" to "<<b<<" takes "<<"-1"<<" knight moves."<<endl;
}
int main()
{
    while(cin>>a>>b)
    {
        memset(visit,0,sizeof(visit));
        int ax=a[0]-'a';
        int ay=a[1]-'1';
        int ax1=b[0]-'a';
        int ay1=b[1]-'1';
        if(a==b)
        {
             cout<<"To get from "<<a<<" to "<<b<<" takes "<<"0"<<" knight moves."<<endl;
        }
        else{
            no=Node(ax1,ay1,0);
        BFS(Node(ax,ay,0));
        }
    }
    return 0;
}