hdoj 2717 Catch That Cow【bfs】
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
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?
for Farmer John to catch the fugitive cow.
#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;
}
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