http://poj.org/problem?id=3278

                                                                                 Catch That Cow
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 47010   Accepted: 14766

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 - 1 or + 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.

数组必须开大点

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h> using namespace std;
struct node
{
int x,ans;
} q[];
int jx[]= {-,};
int n,k;
int map[],v[];
void bfs()
{
struct node t,f;
int e=,s=;
t.x=n;
v[t.x]=;
t.ans=;
q[e++]=t;
while(s<e)
{
t=q[s++];
if(t.x==k)
{
printf("%d\n",t.ans);
break;
}
for(int i=; i<; i++)
{
if(i==)
f.x=t.x*;
else f.x=t.x+jx[i];
if(!v[f.x]&&f.x>=&&f.x<=)
{
f.ans=t.ans+;
q[e++]=f;
v[f.x]=;
}
}
}
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(map,,sizeof(map));
memset(v,,sizeof(v));
bfs();
}
return ;
}

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