Catch That Cow

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

Total Submission(s): 9254 Accepted Submission(s): 2906



[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
Line 1: Two space-separated integers: N and K

[align=left]Output[/align]
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

[align=left]Sample Input[/align]

5 17

[align=left]Sample Output[/align]

4

HintThe 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.

题目要求是让你找直线上从a点到b点的最短距离，没有那么简单，是有条件的，要满足这两个条件：
* 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.//距离是上一段距离的2倍；

程序写好后，提交时发现编译错误，经过查找，发现结构体变量stu now,next;从全局变量变成函数bfs里的变量后在杭电上提交正确

#include<stdio.h>
#include<cstring>
#include<queue>
using namespace std;
int n,k;
int a[200005];
struct stu{
int on;
int step;
};
//struct node
//{
//    int on, step;
//    friend bool operator < (node a, node b)
//    {
//        return a.on > b.on;
//    }
//};
//priority_queue<node>q;
queue<stu>q;
//priority_queue<stu>q;
//int M=100004;
//int IN=1000000;
int bfs()
{
stu now,next;
while(!q.empty())
q.pop();
now.on=n;
now.step=0;
a[now.on]=1;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
//		next.on=now.on;
//		next.step=now.step;
for(int i=0;i<3;i++)
{
if(i==0)
next.on=now.on+1;
if(i==1)
next.on=now.on-1;
if(i==2)
next.on=now.on*2;
next.step=now.step+1;
if(next.on==k)
return next.step;
if(next.on<0||next.on>100000)
continue;
if(!a[next.on])
{
a[next.on]=1;
q.push(next);
}
}
}
}
int main()
{
int h;
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(a,0,sizeof(a));
if(n<k)
{
h=bfs();
printf("%d\n",h);
}
if(n==k)
printf("0\n");
if(n>k)
printf("%d\n",n-k);
}
return 0;
}

再贴一个简单代码：

#include<stdio.h>
#include<cstring>
#include<queue>
using namespace std;
int n,k;
int a[200005];
struct stu{
int on;
int step;
};
int bfs(int n,int k)
{
queue<stu>q;
stu now,next;
now.on=n;
now.step=0;
a[now.on]=1;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
if(now.on==k)
return now.step;
for(int i=0;i<3;i++)
{
if(i==0)
next.on=now.on+1;
if(i==1)
next.on=now.on-1;
if(i==2)
next.on=now.on*2;

if(a[next.on]||next.on<0||next.on>100000)
continue;
a[next.on]=1;
next.step=now.step+1;
q.push(next);
}
}
}
int main()
{
int h;
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(a,0,sizeof(a));

h=bfs(n,k);
printf("%d\n",h);
}
return 0;
}