hdu1372 Knight Moves

最新推荐文章于 2021-07-23 16:46:28 发布

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本文介绍了一种解决国际象棋骑士从一个指定位置到另一个指定位置所需的最少移动次数的问题。通过使用广度优先搜索算法(BFS)，文章提供了一个具体的C++实现方案，有效地解决了旅行骑士问题中的关键挑战。

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Knight Moves

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 33 Accepted Submission(s) : 27

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Problem Description

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
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

The input file will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.

Output

For each test case, print one line saying "To get from xx to yy takes n knight moves.".

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.

因为是按国际象棋里骑士的走法，那么就是8方向的骑士的移动，在走迷宫的基础上把移动的数组里面的数字改一下就可以了。

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int dx[8]={1,2,1,2,-1,-2,-1,-2};
int dy[8]={2,1,-2,-1,2,1,-2,-1};
int ma[9][9];
int sx,sy;
struct po{
int x,y;
int bs;
};
bool check(int x,int y)
{
if(x<1||y<1||x>8||y>8||ma[x][y]==1)
return 0;
return 1;
}
int bfs(int x1,int y1){
queue<po> st;
po tt,hh;
tt.x=x1;
tt.y=y1;
tt.bs=0;
st.push(tt);
ma[x1][y1]=1;
while(!st.empty())
{
tt=st.front();
st.pop();
if(tt.x==sx&&tt.y==sy)
return tt.bs;
for(int i=0;i<8;i++)
{
int xx=tt.x+dx[i];
int yy=tt.y+dy[i];
if(check(xx,yy))
{
hh.x=xx;
hh.y=yy;
hh.bs=tt.bs+1;
ma[xx][yy]=1;
st.push(hh);
}
}
}
}
int main(){
char a[2];
char b[2];
while(cin>>a>>b){
    memset(ma,0,sizeof(ma));
sx=b[0]-'a'+1;
sy=b[1]-'0';
int ans=bfs(a[0]-'a'+1,a[1]-'0');
cout<<"To get from "<<a<<" to "<<b<<" takes "<<ans<<" knight moves."<<endl;
}
return 0;
}