zoj - 1091 - Knight Moves(直接查找法)

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#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;}

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