Uva439：BFS题目总结

 #define _CRT_SECURE_NO_WARNINGS
 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <queue>
 #include <string>
 using namespace std;

 const int maxn =  + ;
 int map[][];
 int cnt;
 int d[][];
 string start, over;
 int dir[][] = { { -,  }, { -,  }, { ,  }, { ,  }, { , - }, { , - }, { -, - }, { -, - } };
　　//棋盘方向通常会用二维数组来表示方向
 struct Node
 {
     int r, c;    // 行,列
     //int dir;     // 方向
     Node(int r = , int c = ) : r(r), c(c) {}
 };

 bool inside(int r, int c)
 {
     if (r <  || r >=  || c <  || c >= ) return false;
     return true;
 }

 //start row, start col, over row, over col
 bool BFS(int sr, int sc, int or, int oc)
 {
     queue<Node> Q;
     memset(d, , sizeof(d));
     Node u(sr, sc);
     Q.push(u);
     while (!Q.empty()) {
         Node u = Q.front(); Q.pop();
         if ((u.r == or) && (u.c == oc)) {      //如果找到，就true;
             return true;
         }
         for (int i = ; i < ; i++) {
             Node v = Node(u.r + dir[i][], u.c + dir[i][]);
             if (inside(v.r, v.c) && d[v.r][v.c] == ) {
                 d[v.r][v.c] = d[u.r][u.c] + ;
                 Q.push(v);
             }
         }
     }
     return false;
 }

 void print_ans(Node u)
 {
     cout << "To get from " << start << " to " << over << " takes " << d[u.r][u.c] << " knight moves.\n";
 }

 int main()
 {
     while (cin >> start >> over) {
         memset(map, , sizeof(map));

         int sr = start[] - 'a', sc = start[] - '' - , or = over[] - 'a', oc = over[] - '' - ;
         BFS(sr, sc, or, oc);
         print_ans(Node(or, oc));
     }
     return ;
 }

总结：通常最短路（最优解）问题会考虑用BFS来进行求解