BFS POJ 3278 Catch That Cow

题目传送门

 /*
     BFS简单题：考虑x-1，x+1，x*2三种情况，bfs队列练练手
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <queue>
 #include <set>
 #include <cmath>
 #include <cstring>
 using namespace std;

 const int MAXN = 1e6 + ;
 const int INF = 0x3f3f3f3f;
 int n, m;
 int d[MAXN];
 bool vis[MAXN];

 void BFS(void)
 {
     memset (vis, , sizeof (vis));
     for (int i=; i<=1e6; ++i)  d[MAXN] = ;

     queue<int> q;
     q.push (n); d[n] = ;   vis[n] = true;

     while (!q.empty ())
     {
         int x = q.front (); q.pop ();
         if (x == m) break;

         int xl = x - ; int xr = x + ; int x2 = x * ;

         if (xl >=  && !vis[xl])
         {
             q.push (xl);    d[xl] = d[x] + ;
             vis[xl] = true;
         }
         if (xr <= 1e6 && !vis[xr])
         {
             q.push (xr);    d[xr] = d[x] + ;
             vis[xr] = true;
         }
         if (x2 <= 1e6 && !vis[x2])
         {
             q.push (x2);    d[x2] = d[x] + ;
             vis[x2] = true;
         }
     }

 }

 int main(void)      //POJ 3278 Catch That Cow
 {
     //freopen ("POJ_3278.in", "r", stdin);

     while (scanf ("%d%d", &n, &m) == )
     {
         BFS ();
         printf ("%d\n", d[m]);
     }

     return ;
 }