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

题意：在一数轴上，农夫位置是N，牛的位置是K，牛不会移动，农夫移动想要抓住牛。农夫有两种移动方式：从X到X-1或X+1，花费一分钟；从X到2*X，花费一分钟，求抓住牛的最小分钟数。

思路：bfs 搜索入门题

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 100001
#define MOD 2520
#define E 1e-12
using namespace std;
bool vis[N];
int dir[2]={-1,1};
struct node
{int x;int step;
}q[N*10];
void bfs(int n,int k)
{int head=1,tail=1;memset(vis,0,sizeof(vis));vis[n]=1;q[tail].x=n;q[tail].step=0;tail++;while(head<tail){int x=q[head].x;int step=q[head].step;int nx;if(x==k){printf("%d\n",step);break;}/*第一种走法*/for(int i=0;i<2;i++){nx=x+dir[i];if(0<=nx&&nx<N&&vis[nx]==0){vis[nx]=1;q[tail].x=nx;q[tail].step=step+1;tail++;}}/*第二种走法*/nx=x*2;if(nx>=0&&nx<N&&vis[nx]==0){vis[nx]=1;q[tail].x=nx;q[tail].step=step+1;tail++;}head++;}
}
int main()
{int n,k;scanf("%d%d",&n,&k);if(k<n){printf("%d",n-k);exit(0);}bfs(n,k);return 0;
}

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Catch That Cow（POJ-3278）

Problem Description

Input

Output

Sample Input

Sample Output

Source Program

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