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2011-05-05 18:19 218 查看

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 22900		Accepted: 7027

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

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.
Source

USACO 2007 Open Silver

简单BFS，注意检查达到的位置是否越界了，re了两次。。。

#include <iostream>
#include <cstring>
#include <queue>

using namespace std;

const int SIZE=210000;

typedef struct
{
int x;
int time;
}Loc;

queue<Loc> lqueue;
bool visit[SIZE];

int main()
{
Loc temp, temp1;
int n,k;
int min_time;

while(cin>>n>>k)
{
memset(visit,0,sizeof(visit));
temp.x=n;
temp.time=0;
visit[temp.x]=true;
lqueue.push(temp);
while(!lqueue.empty())
{
temp=lqueue.front();
lqueue.pop();

if(temp.x==k)//find
{
min_time=temp.time;
break;
}

temp1.time=temp.time+1;
//W op
temp1.x=temp.x+1;
if(!visit[temp1.x] && temp1.x<=100000 && temp1.x>=0)  //检查是否越界
{
visit[temp1.x]=true;
lqueue.push(temp1);
}

temp1.x=temp.x-1;
if(!visit[temp1.x] && temp1.x<=100000 && temp1.x>=0)
{
visit[temp1.x]=true;
lqueue.push(temp1);
}
//T op
temp1.x=2*temp.x;
if(!visit[temp1.x] && temp1.x<=100000 && temp1.x>=0)
{
visit[temp1.x]=true;
lqueue.push(temp1);
}
}
while(!lqueue.empty())
{
lqueue.pop();
}
cout<<min_time<<endl;
}

return 0;
}

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