抓奶牛（广搜）
题目
O - Catch That Cow
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
SubmitStatusPracticePOJ 3278
use MathJax to parse formulas
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 17Sample Output
4Hint
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.
大意
广搜的题目，人在n坐标，牛在m坐标，问如何用最短时间抓到奶牛，每次都可以向前走一步或向后走一步或者飞到2倍的这个点这里
思路
找一个标记数组 记录用了多少分钟 初始化为0 用队列来实现
一开始一直1 100000这组样例不出结果
代码
#include
#include
#include
#include
using namespace std;
int a[100005];
int bfs(int n,int m)
{
queue
int next,head;
q.push(n);
a[n]=1;
while(q.size())
{
head=q.front();
q.pop();
for(int i=0;i<3;i++)
{
if(i==0)
next=head-1;
else if(i==1)
next=head+1;
else
next=head*2;
if(next<0||next>=100001) //一开始一直写的100000导致不对
continue;
if(a[next]==0)
{
q.push(next);
a[next]=a[head]+1;
}
if(next==m)
return a[next];
}
}
}
int main()
{
int n,m,ans,topp;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(a,0,sizeof(a));
if(n>=m)
printf("%d\n",n-m);
else
printf("%d\n",bfs(n,m)-1);
}
return 0;
}