7.12 POJ 3278 A - 广搜 基础

A - 广搜 基础
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
3278

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.

每个节点处有三种走法：+1， -1和*2；

#include<iostream>
#include<queue>
using namespace std;
const int MAXN = 100001;
bool visit[MAXN];
int step[MAXN];
queue<int> que;
int bfs(int n, int k)
{
int head, next;
que.push(n);
visit
= true;
step
== 0;
while(!que.empty())
{
head = que.front();
que.pop();
for(int i = 0; i < 3; i++)
{
if(i == 0) {next = head + 1;}
else if(i == 1){next = head - 1;}
else next = head * 2;
if (next > MAXN || next < 0) continue;
if(!visit[next])
{
que.push(next);
step[next] = step[head] + 1;
visit[next] = true;
}
if(next == k) {return step[next]; }
}
}
}
int main()
{
int n, k;
cin >> n >> k;
cout << bfs(n, k) << endl;
return 0;
}