广搜 象棋格

在一张8*8的象棋盘上，马走日的方式移动，问任意给定的两点pa，pb，问从pa到pb至少需要移动几步。

用 BFS，本题中字符串常量的输入，声明用到了C++中的string类！下为打印样式：

Source:POJ2243

Type:BFS

#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cstdio>#include <queue>#include <stack>#include <algorithm>using namespace std;int dx[] = { 1,  2,  2,  1, -1, -2, -2, -1};int dy[] = { 2,  1, -1, -2, -2, -1,  1,  2};int step[8][8], fa[8][8];//本题没有对路径的要求，加上父亲节点只是为了方便调试bool vis[8][8];inline bool inbound(int x, int y){    return(x>=0&&x<8&&y>=0&&y<8);}inline int encode(int x, int y) { return x * 8 + y; }inline int newPos(int old, int method){    int x = old / 8, y = old % 8;    x += dx[method], y += dy[method];    if (inbound(x, y) )return encode(x, y);    else return -1;}inline int getx(string s) { return int(s[0] - 'a'); }inline int gety(string s) { return int(s[1] - '1'); }int bfs(string _beg, string _tar){    int beg = encode(getx(_beg), gety(_beg));    int tar = encode(getx(_tar), gety(_tar));    memset(step, 0, sizeof(step));    memset(vis, 0, sizeof(vis));    memset(fa, 0, sizeof(fa));    queue<int> q; q.push(beg); vis[beg / 8][beg % 8] = 1;    fa[beg / 8][beg % 8] = beg;    while (!q.empty())    {        int top = q.front(); q.pop();        if (top == tar) break;//跳出循环后，由于step[][]数组被初始化，所以直接是0.        for (int j = 0; j < 8; j++)        {            int tmp = newPos(top, j);            if (tmp != -1 && inbound(tmp/8,tmp%8) && !vis[tmp / 8][tmp % 8])            {                vis[tmp / 8][tmp % 8] = 1;                step[tmp / 8][tmp % 8] = step[top / 8][top % 8] + 1;                fa[tmp / 8][tmp % 8] = top;                q.push(tmp);            }        }    }    return step[tar / 8][tar % 8];}int main(){  string b, t;  while (cin >> b >> t)  printf("To get from %s to %s takes %d knight moves.\n", b.c_str(), t.c_str(), bfs(b, t));  return 0;}