HDU 1372 Knight Moves 题解
Knight Moves
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14125 Accepted Submission(s): 8269
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.
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6
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.
//Author:LanceYu
#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<fstream>
#include<iosfwd>
#include<sstream>
#include<fstream>
#include<cwchar>
#include<iomanip>
#include<ostream>
#include<vector>
#include<cstdlib>
#include<queue>
#include<set>
#include<ctime>
#include<algorithm>
#include<complex>
#include<cmath>
#include<valarray>
#include<bitset>
#include<iterator>
#define ll long long
using namespace std;
const double clf=1e-;
//const double e=2.718281828;
const double PI=3.141592653589793;
const int MMAX=;
//priority_queue<int>p;
//priority_queue<int,vector<int>,greater<int> >pq;
struct node
{
int x,y,step;
};
queue<node> q;
int dir[][]={{-,-},{-,-},{-,},{-,},{,},{,-},{,-},{,}};//马所能够跳的八个方向记录下来
int vis[][];
char temp[][];//定义一个字符串用于输入
int change(char c)//字符转数字
{
switch (c)
{
case 'a':return ;
break;
case 'b':return ;
break;
case 'c':return ;
break;
case 'd':return ;
break;
case 'e':return ;
break;
case 'f':return ;
break;
case 'g':return ;
break;
case 'h':return ;
break;
case '':return ;
break;
case '':return ;
break;
case '':return ;
break;
case '':return ;
break;
case '':return ;
break;
case '':return ;
break;
case '':return ;
break;
case '':return ;
break;
}
}
int bfs(int x,int y,int x1,int y1)
{
while(!q.empty())//队列的初始化,全部清空
q.pop();
int i;
q.push(node{x,y,});
while(!q.empty())
{
node t=q.front();
q.pop();
if(t.x==x1&&t.y==y1)
return t.step;
for(i=;i<;i++)
{
int dx=t.x+dir[i][];
int dy=t.y+dir[i][];
if(dx>=&&dy>=&&dx<&&dy<&&!vis[dx][dy])//基本搜索
{
vis[dx][dy]=;
q.push(node{dx,dy,t.step+});
}
}
}
return ;
}
int main()
{
while(scanf("%s%s",temp[],temp[])!=EOF)
{
memset(vis,,sizeof(vis));
int x=change(temp[][]);
int y=change(temp[][]);
int x1=change(temp[][]);
int y1=change(temp[][]);//确定首尾点
int ans=bfs(x,y,x1,y1);
printf("To get from %s to %s takes %d knight moves.\n",temp[],temp[],ans);//输出
}
return ;
}
2018-11-16 00:03:31 Author:LanceYu
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