UVA 439 Knight Moves

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题意：给国际象棋棋盘的两个坐标，求某个棋子（我也看不懂，规则跟中国象棋里的马一样的）从初始坐标到末坐标的步数，计算最少步数到达目标点并输出，思路就是bfs搜索即可

代码如下

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<cstdlib>#include<iostream>#include<map>#include<queue>#include<vector>#include<stack>#include<set>#include<cctype>#include<istream>#include<ostream>#include<list>#define INF 0x3f3f3f3f#define mem(x) memset(x,0,sizeof(x))#define rep(n) for(int i=1;i<=n;i++)#define rep1(n) for(int j=1;j<=n;j++)const double EPS=1e-6;const double PI=acos(-1.0);typedef long long ll;typedef unsigned long long ull;using namespace std;typedef pair<int,int> p;char c1,c2;int r1,r2,sx,sy,ex,ey;int ans;int d[10][10];int dx[8]= {2,2,-2,-2,1,-1,1,-1};int dy[8]= {1,-1,1,-1,2,2,-2,-2};void bfs(){    rep(8)    rep1(8)    d[i][j]=INF;    d[sx][sy]=0;    queue<p> q;    q.push(p(sx,sy));    while(q.size())    {        p pp=q.front();        q.pop();        int x=pp.first,y=pp.second;        if(x==ex&&y==ey) {ans=d[ex][ey];break;}        if(d[x][y]==INF) continue;        for(int i=0; i<8; i++)        {            int xx=x+dx[i],yy=y+dy[i];            if(d[xx][yy]==INF&&xx>=1&&xx<=8&&yy>=1&&yy<=8) d[xx][yy]=d[x][y]+1;            else continue;            q.push(p(xx,yy));        }    }}int main(){    while(cin>>c1>>r1>>c2>>r2)    {        ans=0;        sx=c1-'a'+1;sy=r1;        ex=c2-'a'+1;ey=r2;        bfs();        printf("To get from %c%d to %c%d takes %d knight moves.\n",c1,r1,c2,r2,ans);    }}

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