POJ - 3278 Catch That Cow 简单搜索

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.

这道题是关于最短步数的，首先就要想到广搜，既然想到了广搜就可以写代码了。

#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#define Swap(a,b) a^=b^=a^=b
#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define speed ios_base::sync_with_stdio(0);//Çв»¿ÉÓÃscnaf£»
#define Max(a,b) a>b?a:b
#define Min(a,b) a<b?a:b
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e6+10;
const double esp=1e-9;
int m,n,x,y;
int cnt,mini=INF;
bool flag[maxn];
int bfs(int x);
int main()
{
    cin>>m>>n;
    cout<<bfs(m)<<endl;
}
int bfs(int x)
{
    queue<pair<int,int> > t;
    t.push(make_pair(x,0));
    while(!t.empty())
    {
        pair<int,int>w=t.front();
        if(w.first==n)break;
        t.pop();
        w.first++,w.second++;
        if(w.first<0||w.first>100000||flag[w.first]);
        else t.push(w),flag[w.first]=1;
        w.first-=2;
        if(w.first<0||w.first>100000||flag[w.first]);
        else t.push(w),flag[w.first]=1;
        w.first++;
        w.first*=2;
        if(w.first<0||w.first>100000||flag[w.first]);
        else t.push(w),flag[w.first]=1;
    }
    return t.front().second;
}