Catch That Cow
Time Limit: 2000 ms Memory Limit: 65536 KiB
Problem Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting. * Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute * Teleporting: FJ can move from any point X to the point 2 × X in a single minute. If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Line 1: Two space-separated integers: N and K
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input
5 17
Sample Output
4
提示：如果你要做POJ的题，你要注意一点，当用C++的时候，不要用到bits/stdc++.h万能头文件，因为POJ判题中不存在此头文件，HDU部分判题也不包含此头文件
此题主要就是用到广度遍历BFS，遍历能到达牛的位置的点
代码实现如下（g++）：
# include <cstdio> # include <string.h> # include <queue> using namespace std; int n,m,ans; int num[100010]; int main(void) { while(~scanf("%d %d",&n,&m)) { queue<int>t; t.push(n); num[n]=1; while(!t.empty()) { int c=t.front(); if(c==m) { ans=num[c]; break; } t.pop(); if(c-1>=0&&!num[c-1])//如果c-1没有被遍历过，将其入队 { num[c-1]=num[c]+1; t.push(c-1); } if(c+1<=100000&&!num[c+1])//如果c+1没有被遍历过，将其入队 { num[c+1]=num[c]+1; t.push(c+1); } if(c*2<=100000&&!num[c*2])//如果c*2没有被遍历过，将其入队 { num[c*2]=num[c]+1; t.push(c*2); } } printf("%d ",ans-1); } return 0; } /*************************************************** Result: Accepted Take time: 0ms Take Memory: 892KB ****************************************************/