题干：

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

Hint

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.

解题报告

   这题用bfs去递推（其实也相当于dp啦，可以理解成  只是不是按照for循环遍历的顺序去dp而已）

AC代码：

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
bool vis[300000 + 5];
int n,k;
struct Node {int pos,step;Node(){}Node(int pos,int step):pos(pos),step(step){}
};
int bfs(int x) {queue<Node> q;memset(vis,0,sizeof vis);q.push(Node(x,0));vis[x]=1;while(!q.empty()) {Node cur = q.front();q.pop();if(cur.pos == k) return cur.step;if( cur.pos+1<= 300000 && cur.pos+1 >= 0&&!vis[cur.pos+1] ){vis[cur.pos+1]=1;q.push(Node(cur.pos+1,cur.step+1));}if(cur.pos-1<= 300000&& cur.pos-1  >= 0 &&!vis[cur.pos-1]  ) {vis[cur.pos-1]=1;q.push(Node(cur.pos-1,cur.step+1));}if((cur.pos<<1) <= 300000 && (cur.pos<<1) >=0 &&!vis[cur.pos<<1]  ) {vis[cur.pos<<1]=1 ;q.push(Node(cur.pos<<1,cur.step+1)); 	}}return 0;
}
int main()
{while(~scanf("%d%d",&n,&k)) {printf("%d\n",bfs(n));}return 0;
}

错误代码：（dp）

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
ll dp[3000000 +5];
inline ll min3(ll i,ll j,ll k) {if(i < j) {return i < k ? i : k;}if(j < k) return j;else return k;
}
int main()
{int n,k;while(~scanf("%d%d",&n,&k)) {memset(dp,INF,sizeof dp);for(int i = 1; i<=n; i++) {dp[i] = min(dp[i],(ll)n-(ll)i);dp[i*2] = min(dp[i*2],dp[i]+1);}if(k <= n) {printf("%lld\n",dp[k]);continue;}for(int i = n+1; i<=(k<<1); i++) {dp[i] = min3(dp[i],dp[i-1]+ 1,dp[i+1] + 1) ;dp[i*2] = min(dp[i*2],dp[i]+1);}printf("%lld\n",dp[k]);}return 0;
}

总结：

   这题有个很大的坑啊！！我刚开始不管数组开多大都是re，后来发现，，，数组不需要开多大，因为他有可能是负数！！所以需要判断是否越界。

   其二，开30W的数组就够了，不需要300W，但是有一点要注意，判断条件中，必须判断范围在前，判断vis数组在后。

      即，判断越界在前，判断其他在后。这一点非常重要！我立马想到了一般我写地图类dfs中的tx，ty也没大注意先判断越界还是先判断maze是否满足还是先判断vis是否标记过，这是因为！！！我的地图都是从1开始读入的，所以就算是先判断别的在判断txty是否越界，也不会出现re的情况，这一点是我今天才注意到的、、惭愧惭愧、、、if中判断顺序有的时候也是有区别的！！！有可能会导致越界！！