POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 62404 Accepted: 19533

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

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.

题目大意 有一个人在找自己的奶牛，他和奶牛在一条直线的路上，他的位置是N奶牛的位置是K，他每分钟可以进行一个操作，1.向前走一步，2.向后走一步，3.往前面跳到自己当前位置两倍的位置，问最少多分钟找到自己的奶牛。

思路 裸BFS

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;#define inf 0xffffff#define maxn 100005int Map[maxn];int q[maxn],t1,t2;int main(){    int s,e,ans;    while(cin>>s>>e){        memset(Map,0,sizeof(Map));        t1 = t2 = 0;        q[t2++] = s;        Map[s] = 1;        while(s != e){            s = q[t1++];            if(s*2 < maxn && !Map[s*2]) {Map[s*2] = Map[s] + 1; q[t2++] = s * 2;}            if(s+1 < maxn && !Map[s+1]) {Map[s+1] = Map[s] + 1; q[t2++] = s + 1;}            if(s-1 >= 0 && !Map[s-1]) {Map[s-1] = Map[s] + 1; q[t2++] = s - 1;}        }        cout<<Map[e] - 1<<endl;    }    return 0;}