杭电 HOJ 1372 Knight Moves 解题报告

马走棋盘，搜索。刚开始想打表，发现棋子在边界的时候很难处理，比如a1到b2的话2步是不行的。

用队列，时效差一点点，46MS，代码如下：

#include<iostream>
#include<deque>
using namespace std;

struct Point
{
char x,y,num;
};

int main()
{
int a,b,i,j;
Point x,y;
deque<Point> q;
char str[2],str2[2];
char visit[8][8];
while(cin>>str>>str2)
{
memset(visit,0,sizeof(visit));
x.x=str[0]-'a';
x.y=str[1]-'1';
x.num=0;
q.clear();
q.push_back(x);
visit[x.x][x.y]=1;
a=str2[0]-'a';
b=str2[1]-'1';

while(!(x.x==a&&x.y==b))
{
y.num=x.num+1;
if(x.x>=2 && x.y>=1 && !visit[y.x=x.x-2][y.y=x.y-1])
{
q.push_back(y);
visit[y.x][y.y]=1;
}
if(x.x<6 && x.y>=1 && !visit[y.x=x.x+2][y.y=x.y-1])
{
q.push_back(y);
visit[y.x][y.y]=1;
}
if(x.x<6 && x.y<7 && !visit[y.x=x.x+2][y.y=x.y+1])
{
q.push_back(y);
visit[y.x][y.y]=1;
}
if(x.x>=2 && x.y<7 && !visit[y.x=x.x-2][y.y=x.y+1])
{
q.push_back(y);
visit[y.x][y.y]=1;
}
if(x.x>=1 && x.y>=2 && !visit[y.x=x.x-1][y.y=x.y-2])
{
q.push_back(y);
visit[y.x][y.y]=1;
}
if(x.x>=1 && x.y<6 && !visit[y.x=x.x-1][y.y=x.y+2])
{
q.push_back(y);
visit[y.x][y.y]=1;
}
if(x.x<7 && x.y<6 && !visit[y.x=x.x+1][y.y=x.y+2])
{
q.push_back(y);
visit[y.x][y.y]=1;
}
if(x.x<7 && x.y>=2 && !visit[y.x=x.x+1][y.y=x.y-2])
{
q.push_back(y);
visit[y.x][y.y]=1;
}

q.pop_front();
x=q.front();
}
cout<<"To get from "<<str<<" to "<<str2<<" takes "<<(int)x.num<<" knight moves."<<endl;
}
}

忽然发现自己曾经过了这题，代码是网上找的，也贴上吧~思路是一样的，循环部分比我的简洁。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<iomanip>
#include<string>
#include<algorithm>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
class Cordinate
{public:int x,y;};
using namespace std;
int dir[8][2]={1,2,1,-2,2,1,2,-1,-1,2,-1,-2,-2,1,-2,-1},chess[10][10];
bool legal(Cordinate po)
{if(po.x>=1&&po.x<=8&&po.y>=1&&po.y<=8)return true;return false;}
int main()
{

char a[3],b[3];Cordinate po,pos;int i;
while(cin>>a>>b)
{
queue<Cordinate>q;
memset(chess,-1,sizeof(chess));
pos.x=a[0]-'a'+1;pos.y=a[1]-'0';
chess[pos.x][pos.y]=0;
q.push(pos);
while(!q.empty())
{
po=q.front();q.pop();
for(i=0;i<8;i++)
{
pos.x=dir[i][0]+po.x;pos.y=dir[i][1]+po.y;
if(legal(pos)&&chess[pos.x][pos.y]==-1) {q.push(pos);chess[pos.x][pos.y]=chess[po.x][po.y]+1;}
}
}
cout<<"To get from "<<a<<" to "<<b<<" takes "<<chess[b[0]-'a'+1][b[1]-'0']<<" knight moves."<<endl;
}
return 0;
}