HDU2717Catch That Cow

简单的搜索题目，，渣渣在这里给大家展示一下

题目大意：简单的搜索题，即有三种走法，最后要求抓到那头奶牛

第一种走法：向前1格

第二种走法：走到是本位置2倍的位置

第三种走法：向后一格

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10312 Accepted Submission(s): 3221

Problem 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

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.

代码：

include

include

include

using namespace std;

int a[100005],book[100005];

int bfs(int x,int y)

{

a[x]=0;

book[x]=0;

int num,sum;//num为所处的位置，sum为可以到达的位置

memset(a,0,sizeof(a));

memset(book,0,sizeof(book));

queueq;

q.push(x);

while(!q.empty())

{

num=q.front();

q.pop();

if(num==y)

{

break;

}

for(int i=1;i<=3;i++)

{

if(i==1)

{

sum=num+1;

}

if(i==2)

{

sum=num*2;

}

if(i==3)

{

sum=num-1;

}

if(sum>=0&&sum<=100000&&book[sum]==0)

{

book[sum]=1;

a[sum]=a[num]+1;

q.push(sum);

}

}

}

return a[y];

}

int main()

{

int m,n;

while(~scanf(“%d %d”,&m,&n))

{

printf(“%d\n”,bfs(m,n));

}

return 0;

}