Chess

链接：http://codeforces.com/problemset/problem/38/B

题目：

Two chess pieces, a rook and a knight, stand on a standard chessboard
8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.

Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.

题意：8*8棋盘，有一个马一个车，求再放一个马有多少种方法。

分析：数据很小，暴力把不可能的点去掉即可。注意放第二个马的时候不能踩到车。

看自己之前写的代码好蠢，这里马的8个走位可以用数组表示，类似bfs统一用一条语句判断越界，其他文章可能会有提到。

题解：

#include<iostream>
using namespace std;

bool board[10][10];

void killrook(int m,int n)
{
for(int i=0;i<8;i++)
{
board[m][i]=board[i][n]=true;
}
}
void killknight(int m,int n)
{
board[m][n]=true;
if(m-2>=0)
{
if(n-1>=0)
board[m-2][n-1]=true;
if(n+1<8)
board[m-2][n+1]=true;
}
if(m+2<8)
{
if(n-1>=0)
board[m+2][n-1]=true;
if(n+1<8)
board[m+2][n+1]=true;
}
if(n-2>=0)
{
if(m-1>=0)
board[m-1][n-2]=true;
if(m+1<8)
board[m+1][n-2]=true;
}
if(n+2<8)
{
if(m-1>=0)
board[m-1][n+2]=true;
if(m+1<8)
board[m+1][n+2]=true;
}
}
int main()
{
//freopen("in.txt","r",stdin);
char c;
int i;
while(~scanf("%c%d",&c,&i))
{
int ans=0;
memset(board,0,sizeof(board));
killrook(i-1,c-'a');
killknight(i-1,c-'a');
getchar();
scanf("%c%d",&c,&i);
getchar();
killknight(i-1,c-'a');
for(int j=0;j<8;j++)
for(int k=0;k<8;k++)
{
if(board[j][k]==false)
ans++;
}
printf("%d\n",ans);
}
return 0;
}