Chess

Description
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.

Input
The first input line contains the description of the rook’s position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight’s position in a similar way. It is guaranteed that their positions do not coincide.

Output
Print a single number which is the required number of ways.

Sample Input
Input
a1
b2
Output
44
Input
a8
d4
Output
38
这是一道与下象棋有关的题目，题意是给你一个车和马的坐标，然后再给你一个马，让你判断最后三者都不会被吃掉的可能性，注意的是三者是三个独立的个体。
这道题要知道一点下象棋的规则；马走“日”字， 车可以直线随便走，然后判断车和马的八个方向和车的行和列即可。
我本身不会下象棋，这让我有理由要学会象棋了。
给一下代码：

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;int tp[10][10];int dir[8][2] = {2, 1, 2, -1, -2, 1, -2, -1, 1, 2, 1, -2, -1, 2, -1, -2};void init(){    for(int i = 0; i <= 8; i++)    {        for(int j = 0; j <= 8; j++)        {            tp[i][j] = 0;        }    }}int check(int x, int y){    if(x >= 1 && x <= 8 && y >= 1 && y <= 8) return 1;    return 0;}int main(){    char a[4], b[4];    int x1, y1, x2, y2, sum;    while(~scanf("%s %s", a, b))    {        init();        sum = 0;        x1 = a[0] - 'a' + 1;        y1 = a[1] - '0';        x2 = b[0] - 'a' + 1;        y2 = b[1] - '0';        tp[x1][y1] = tp[x2][y2] = 1;        for(int k = 0; k < 8; k++)        {            int x0 = x2 + dir[k][0];            int y0 = y2 + dir[k][1];            if(check(x0, y0)) tp[x0][y0] = 2;        }        for(int k = 0; k < 8; k++)        {            int x0 = x1 + dir[k][0];            int y0 = y1 + dir[k][1];            if(check(x0, y0)) tp[x0][y0] = 2;        }        for(int i = 1; i <= 8; i++)        {            tp[x1][i] = 3;            tp[i][y1] = 3;        }        for(int i = 1; i <= 8; i++)        {            for(int j = 1; j <= 8; j++)            {//                printf("%d ", tp[i][j]);                if(!tp[i][j]) sum++;            }//            puts("");        }        printf("%d\n", sum );    }    return 0;}