1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <queue>
 6 #include <string>
 7 using namespace std;
 8 
 9 const int maxn = 500 + 20;
10 int map[9][9]; 
11 int cnt;
12 int d[9][9];
13 string start, over;
14 int dir[8][2] = { { -2, 1 }, { -1, 2 }, { 1, 2 }, { 2, 1 }, { 2, -1 }, { 1, -2 }, { -1, -2 }, { -2, -1 } };
　　//棋盘方向通常会用二维数组来表示方向
15 struct Node
16 {
17     int r, c;    // 行,列
18     //int dir;     // 方向
19     Node(int r = 0, int c = 0) : r(r), c(c) {}
20 };
21 
22 bool inside(int r, int c)
23 {
24     if (r < 0 || r >= 8 || c < 0 || c >= 8) return false;
25     return true;
26 }
27 
28 //start row, start col, over row, over col
29 bool BFS(int sr, int sc, int or, int oc)
30 {
31     queue<Node> Q;
32     memset(d, 0, sizeof(d));
33     Node u(sr, sc);
34     Q.push(u);
35     while (!Q.empty()) {
36         Node u = Q.front(); Q.pop();
37         if ((u.r == or) && (u.c == oc)) {      //如果找到，就true;
38             return true;
39         }
40         for (int i = 0; i < 8; i++) {
41             Node v = Node(u.r + dir[i][0], u.c + dir[i][1]);
42             if (inside(v.r, v.c) && d[v.r][v.c] == 0) {
43                 d[v.r][v.c] = d[u.r][u.c] + 1;
44                 Q.push(v);
45             }
46         }
47     }
48     return false;
49 }
50 
51 void print_ans(Node u)
52 {
53     cout << "To get from " << start << " to " << over << " takes " << d[u.r][u.c] << " knight moves.\n";
54 }
55 
56 int main()
57 {
58     while (cin >> start >> over) {
59         memset(map, 0, sizeof(map));
60         
61         int sr = start[0] - 'a', sc = start[1] - '0' - 1, or = over[0] - 'a', oc = over[1] - '0' - 1;
62         BFS(sr, sc, or, oc);
63         print_ans(Node(or, oc));
64     }
65     return 0;
66 }

总结：通常最短路（最优解）问题会考虑用BFS来进行求解