转载请注明出处：https://blog.csdn.net/Mercury_Lc/article/details/82693928 作者：Mercury_Lc
题目链接
题解：给你x、y，x可以加1、减1、或者变成2*x，问通过最少的次数来让x等于y，这是最基础的bfs，就是把x通过一次的 1、-1、*2得到的数都放到队列里面，再把这些通过一次操作得到的数进行相同的操作 1、-1、*2，因为用个结构体来存放这个数是第几次操作得到的，所以只要一旦发现这个数，一定是通过最小的次数得到的。
代码语言：javascript复制#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
const int maxn = 1e6 10;
int vis[maxn];
struct node
{
int data;
int step;
} w,l;
void bfs(int n,int k)
{
memset(vis,0,sizeof(vis));
vis[n] = 1;
queue<node>q;
w.data = n;
w.step = 0;
q.push(w);
while(!q.empty())
{
w = q.front();
q.pop();
if(w.data == k)
{
printf("%dn",w.step);
return ;
}
if(w.data 1 <= maxn && !vis[w.data 1])
{
l = w;
l.data = 1;
l.step ;
q.push(l);
vis[l.data] = 1;
}
if(w.data - 1 <= maxn && w.data - 1 >= 0&& !vis[w.data - 1])
{
l = w;
l.data -= 1;
l.step ;
q.push(l);
vis[l.data] = 1;
}
if(w.data * 2 <= maxn && !vis[w.data * 2])
{
l =w;
l.step ;
l.data *= 2;
q.push(l);
vis[l.data] = 1;
}
}
return ;
}
int main()
{
int n,k;
while(~scanf("%d %d",&n,&k))
{
if(n > k)
printf("%dn",n - k);
else
bfs(n, k);
}
return 0;
}Problem 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 17Sample Output4Hint The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
