poj3278——Catch That Cow（BFS，剪枝）

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

n只有三种变换情况，+1，-1，x2，求n最少能经过多少步成为k。

乍看很简单，对三种情况BFS就行了，但这个数据很大，容易TLE或者RE。

我第一次TLE，之后就设置了一个观察数组，如果某次BFS时又碰到了这个数，这时的次数肯定不是最小的。

第二次RE，又考虑了n大于k的情况，这时候肯定只能一直-1，所以特判。又x2的操作容易超数组，所以如果某次x2超过了目标的两倍，那肯定是没用的，剪掉。到达负数也不行，剪掉。最后就过了

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <set>
#include <vector>
#include <iomanip>
#include <stack>
#include <map>
#include <queue>
#define MAXN 100010
#define mod 100010
#define INF 0x3f3f3f3f
using namespace std;
int n,k,ans;
int vis[MAXN*10];
struct Node
{
int num;
int step;
};
void bfs(int n)
{
queue<Node> q;
Node start,tmp1,tmp2;
start.num=n;
start.step=0;
q.push(start);
while(!q.empty())
{
tmp1=q.front();
q.pop();
if(tmp1.num==k)
{
ans=min(ans,tmp1.step);
return;
}
if(tmp1.num-1>=0)
{
tmp2.num=tmp1.num-1;
tmp2.step=tmp1.step+1;
if(!vis[tmp2.num])
{
vis[tmp2.num]=1;
q.push(tmp2);
}
}
tmp2.num=tmp1.num+1;
tmp2.step=tmp1.step+1;
if(!vis[tmp2.num])
{
vis[tmp2.num]=1;
q.push(tmp2);
}
if(tmp1.num*2<k*2)
{
tmp2.num=tmp1.num*2;
tmp2.step=tmp1.step+1;
if(!vis[tmp2.num])
{
vis[tmp2.num]=1;
q.push(tmp2);
}
}
}
}
int main()
{
ios::sync_with_stdio(false);
while(cin>>n>>k)
{
ans=INF;
if(n>=k)
{
cout<<n-k<<endl;
continue;
}
memset(vis,0,sizeof(vis));
bfs(n);
cout<<ans<<endl;
}
return 0;
}