Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9276 Accepted Submission(s):
2907
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.
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.
for Farmer John to catch the fugitive cow.
Sample Input
5 17
Sample Output
4
题意：人在n位置处，牛在k位置处，牛不动问人最短多少分钟可以捉到牛，人有两种移动方法
1、一分钟移动一步n-1或者n+1
2、一分钟移动2*n步；
#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX 1000100
using namespace std;
int n,m;
int vis[MAX];
struct node
{
int x;
int step;
friend bool operator < (node a,node b)
{
return a.step>b.step;
}
};
int judge(int x)
{
if(x < 0||x > MAX||vis[x])
return 0;
return 1;
}
void bfs(int n,int k)
{
int i,j;
priority_queue<node>q;
node beg,end;
beg.x=n;
beg.step=0;
q.push(beg);
vis[n]=1;
while(!q.empty())
{
beg=q.top();
q.pop();
if(beg.x==k)
{
printf("%d\n",beg.step);
return ;
}
end.x=beg.x-1;
if(judge(end.x))
{
vis[end.x]=1;
end.step=beg.step+1;
q.push(end);
}
end.x=beg.x+1;
if(judge(end.x))
{
vis[end.x]=1;
end.step=beg.step+1;
q.push(end);
}
end.x=beg.x*2;
if(judge(end.x))
{
vis[end.x]=1;
end.step=beg.step+1;
q.push(end);
}
}
}
int main() {
int i;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(vis,0,sizeof(vis));
bfs(n,m);
}
return 0;
}