zoj - 1091 - Knight Moves(直接查找法)

#include <iostream>
#include <string>
using namespace std;
int main()
{int f[8][8] ={
   {0, 3, 2, 3, 2, 3, 4, 5},     //因为是8*8的方阵，不是很多数，所以枚举了所有的情况，就是从点(0, 0)到(x, y)最少要几步{3, 2, 1, 2, 3, 4, 3, 4},{2, 1, 4, 3, 2, 3, 4, 5},{3, 2, 3, 2, 3, 4, 3, 4},{2, 3, 2, 3, 4, 3, 4, 5},{3, 4, 3, 4, 3, 4, 5, 4},{4, 3, 4, 3, 4, 5, 4, 5},{5, 4, 5, 4, 5, 4, 5, 6}};string a, b;while(cin>>a>>b){if((a == "a1" && b == "b2") || (a == "b2" && b == "a1")     //小心一种特殊情况：边缘情况|| (a == "h1" && b == "g2") || (a == "g2" && b == "h1")|| (a == "a8" && b == "b7") || (a == "b7" && b == "a8")|| (a == "h8" && b == "g7") || (a == "g7" && b == "h8"))cout<<"To get from "<<a<<" to "<<b<<" takes "<<4<<" knight moves."<<endl;else{int x = (int)(a[0] - b[0]);     //计算差距，直接输出int y = (int)(a[1] - b[1]);x = x >=0 ? x : -x;y = y >=0 ? y : -y;cout<<"To get from "<<a<<" to "<<b<<" takes "<<f[x][y]<<" knight moves."<<endl;}}return 0;
}