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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
Output
Sample Input
5 17
Sample Output
4
Hint
简单题，不说，直接代码。。。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define MAX 100005
#define INF 0x3f3f3f3f
#define LL long long
#define pii pair<int,int>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rd3(x,y,z) scanf("%d%d%d",&x,&y,&z)
///map<int,int>mmap;
///map<int,int >::iterator it;
using namespace std;
struct Pos
{
int x;
int step;
Pos() {}
Pos(int x,int step)
{
this->x=x,this->step=step;
}
};
bool vis[MAX];
int main ()
{
int m,n;
while(~rd2(m,n))
{
memset(vis,0,sizeof(vis));
if(m>n)
{
printf("%d\n",m-n);
continue;
}
queue<Pos> que ;
que.push(Pos(m,0));
while(!que.empty())
{
Pos temp=que.front();
if(temp.x==n)
{
printf("%d\n",temp.step);
break;
}
que.pop();
if( temp.x-1>=0 && !vis[temp.x-1] )
{
que.push(Pos(temp.x-1,temp.step+1));
vis[temp.x-1]=true;
}
if( temp.x+1<=MAX-5 && !vis[temp.x+1] )
{
que.push(Pos(temp.x+1,temp.step+1));
vis[temp.x+1]=true;
}
if( temp.x<<1<=MAX-5 && !vis[temp.x<<1])
{
que.push(Pos(temp.x*2,temp.step+1));
vis[temp.x<<1]=true;
}
}
}
return 0;
}
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原文地址：http://blog.csdn.net/u014665013/article/details/51329350
