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?
* 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.
之所以把这个水题贴上，是为了警戒自己，数组超限与范围判断的时候，应该先进行范围判断，然后进行数组操作！
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
int vis[300010];
int bfs(int x,int ans){
queue<int>q;
queue<int>num;
q.push(x);
num.push(0);
while(!q.empty()){
int xx = q.front();
int nn = num.front();
num.pop();
q.pop();
if(xx == ans)return nn;
if(xx>=0&&vis[xx-1] == 0){
q.push(xx-1);
num.push(nn+1);
vis[xx-1] = 1;
}
if(xx+1<=ans&&vis[xx+1] == 0){
q.push(xx+1);
num.push(nn+1);
vis[xx+1] = 1;
}
if(2*xx <=ans*2&&vis[2*xx] == 0){
q.push(2*xx);
num.push(nn+1);
vis[2*xx] = 1;
}
}
}
int main(){
int n,m;
while(~scanf("%d%d",&n,&m)){
memset(vis,0,sizeof(vis));
vis[n]=1;
printf("%d
",bfs(n,m));
}
}