POJ Knight Moves 2243 x
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 13974 | Accepted: 7797 |
Description
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Input
Output
Sample Input
e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6
Sample Output
To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.
Source
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring> using namespace std; char s[];
bool v[][];
int ex,ey,sx,sy,ans;
int dx[]={,,-,-,-,-,,};
int dy[]={-,-,-,-,,,,};//八个方向 struct node
{
int x,y,step;
}cur,nxt; queue<node>q; void bfs()
{
if(ex==sx&&ey==sy) //特判起点等于终点 ,找到后进行输出就一定是最小的
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",char(ex+'a'-),ey,char(sx+'a'-),sy,);
return;//格式
}
while(!q.empty()) q.pop(); // 多组数据初始化
memset(v,,sizeof(v)); // 同上
cur.x=ex,cur.y=ey; cur.step=; //起点
v[ex][ey]=true; //不要漏了标记起点
q.push(cur);
while(!q.empty())
{
cur=q.front();
q.pop(); //不要漏了当前出队
//v[cur.x][cur.y]=false; 出队,清除标记,是否需要?不需要,为什么?
for(int i=;i<;i++) //八方位搜索
{
int xx=cur.x+dx[i],yy=cur.y+dy[i];
if(xx>&&xx<=&&yy>&&yy<=&&!v[xx][yy])
{
if(xx==sx&&yy==sy) //找到了,第一个找到的一定就是最近的,why?
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",char(ex+'a'-),ey,char(sx+'a'-),sy,cur.step+);
return ;
}
nxt.x=xx, nxt.y=yy; nxt.step=cur.step+;
v[nxt.x][nxt.y]=true;
q.push(nxt); //扩展出的状态入队
}
}
}
} int main()
{
while(scanf("%s",s)!=EOF) //注意输入,scanf读到空格
{
ex=s[]-'a'+; ey=s[]-'';
scanf("%s",s);
sx=s[]-'a'+; sy=s[]-'';
bfs();
}
}
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