POJ 3278 Catch That Cow(DFS)

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 pointsX - 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.

题解：分情况，设f(n,t) 是 n 到 t 的最短时间。
1.如果n>k，直接输出n-k；

2.如果n=0，那么走n-1和走2*n都是无意义的，只能走n+1，此时n先走到1，故f(0,k)=(1,k)+1，且令n=1往下操作；

3.如果k是偶数 ，a情况：如果k/2>n,f(n，k) = f(n，k/2)+1；
                           b情况：如果k/2<n,f(n，k) = min( f(n，k/2)+1，k-n(直接一步步过去) );

4.如果k是奇数，那么必定是f(n,k) = min(f(n，k+1)，f(n，k-1))+1;

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int  n,m;int dfs(int cur,int target){    if(cur>=target)        return cur-target;    if(target%2==0)    {        if(cur<target/2)            return dfs(cur,target/2)+1;        else if(cur==target/2)            return 1;        else            return min(1+cur-target/2,target-cur);    }    else    {        return min(dfs(cur,target+1),dfs(cur,target-1))+1;    }}int main(){    int flag=0;    scanf("%d %d",&n,&m);    if(n==0)        flag=1;    printf("%d\n",dfs(n+flag,m)+flag);    return 0;}