461

# 题目

## 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.

5 17

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.

# 代码

/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！

*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o<n;o++)

const int maxn = 100005;

int BFS(int s,int v) {
if(s == v)
return 0;
if(s > v)
return s - v;

queue<int> Q;
bool visited[maxn];
memset(visited,false,sizeof(visited));
int dis[maxn];
memset(dis,0,sizeof(dis));

Q.push(s);
visited[s] = true;
while(!Q.empty()) {
int th = Q.front();
Q.pop();

//达到终点
if(th == v)
break;

//拓展节点

define push \
if(next > maxn || next <= 0) \
continue;\
if(!visited[next]) {\
Q.push(next);\
visited[next] = true;\
dis[next] = dis[th] + 1;\
}\

int next;
next = th + 1;
push;
next = th - 1;
push;
next = th * 2;
push;
}

if(dis[v])
return dis[v];
else
return -1;
}

bool Do() {
int s,v;
if(scanf("%d%d",&s,&v)==EOF)
return false;
printf("%d\n",BFS(s,v));
return true;
}

int main() {
while(Do());
return 0;
}

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