POJ 2243 象棋

 一个8*8的中国象棋棋牌,给你两个坐标,问你马从起点走到终点最少需要几步.(马可以朝4个方向8种走法,只能走日字,具体见代码)

 

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int maxn=10;
int r1,c1,r2,c2;
struct Node
{
    int r,c;
    Node(int r,int c):r(r),c(c){}
};
int vis[maxn][maxn];
int dist[maxn][maxn];
queue<Node> Q;
int dr[]={-2,-2,-1,-1,1,1,2,2};
int dc[]={-1,1,2,-2,2,-2,1,-1};
int BFS()
{
    if(r1==r2&&c1==c2) return 0;
    while(!Q.empty()) Q.pop();
    memset(vis,0,sizeof(vis));
    vis[r1][c1]=1;
    dist[r1][c1]=0;
    Q.push(Node(r1,c1));
    while(!Q.empty())
    {
        Node node=Q.front();Q.pop();
        int r=node.r,c=node.c;
        for(int d=0;d<8;d++)
        {
            int nr=r+dr[d];
            int nc=c+dc[d];
            if(nr>=0&&nr<8 && nc>=0 &&nc<8 &&vis[nr][nc]==0)
            {
                if(nr==r2&&nc==c2) return dist[r][c]+1;
                dist[nr][nc]=1+dist[r][c];
                Q.push(Node(nr,nc));
                vis[nr][nc]=1;
            }
        }
    }
    return -1;
}
int main()
{
    char str1[10],str2[10];
    while(scanf("%s%s",str1,str2)==2)
    {
        r1=str1[1]-'1';
        c1=str1[0]-'a';
        r2=str2[1]-'1';
        c2=str2[0]-'a';
        printf("To get from %s to %s takes %d knight moves.\n",str1,str2,BFS());
    }
    return 0;
}