BFS POJ 3278 Catch That Cow

 

题目传送门

 1 /*  2  BFS简单题：考虑x-1，x+1，x*2三种情况，bfs队列练练手  3 */  4 #include <cstdio>  5 #include <iostream>  6 #include <algorithm>  7 #include <map>  8 #include <queue>  9 #include <set> 10 #include <cmath> 11 #include <cstring> 12 using namespace std; 13 14 const int MAXN = 1e6 + 10; 15 const int INF = 0x3f3f3f3f; 16 int n, m; 17 int d[MAXN]; 18 bool vis[MAXN]; 19 20 void BFS(void) 21 { 22 memset (vis, 0, sizeof (vis)); 23 for (int i=0; i<=1e6; ++i) d[MAXN] = 0; 24 25 queue<int> q; 26 q.push (n); d[n] = 0; vis[n] = true; 27 28 while (!q.empty ()) 29  { 30 int x = q.front (); q.pop (); 31 if (x == m) break; 32 33 int xl = x - 1; int xr = x + 1; int x2 = x * 2; 34 35 if (xl >= 0 && !vis[xl]) 36  { 37 q.push (xl); d[xl] = d[x] + 1; 38 vis[xl] = true; 39  } 40 if (xr <= 1e6 && !vis[xr]) 41  { 42 q.push (xr); d[xr] = d[x] + 1; 43 vis[xr] = true; 44  } 45 if (x2 <= 1e6 && !vis[x2]) 46  { 47 q.push (x2); d[x2] = d[x] + 1; 48 vis[x2] = true; 49  } 50  } 51 52 } 53 54 int main(void) //POJ 3278 Catch That Cow 55 { 56 //freopen ("POJ_3278.in", "r", stdin); 57 58 while (scanf ("%d%d", &n, &m) == 2) 59  { 60  BFS (); 61 printf ("%d\n", d[m]); 62  } 63 64 return 0; 65 }