HDU 2717 Catch That Cow (bfs)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2717
Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12615 Accepted Submission(s):
3902
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.
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.
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
using namespace std;
int f[]; //记录步数
void bfs(int n,int k)
{
int a[]; //位置数组
memset(f,,sizeof(f));
queue <int > q;
q.push(n);
f[q.front()] = ;
if (n == k)
return ;
while (!q.empty())
{
int t = q.front();
int x = f[t];
a[] = t-; //三个位置
a[] = t+;
a[] = t*;
for (int i = ; i < ; i ++)
{
if (a[i] >= && a[i] < && f[a[i]]==) //注意这里的边界值
{
q.push(a[i]);
f[a[i]] = x+; //在上一步的基础上加1
}
if (a[i] == k)
return ;
}
q.pop();
}
}
int main ()
{
int i;
int n,k;
while (~scanf("%d%d",&n,&k))
{
dfs(n,k);
printf("%d\n",f[k]-); //减去农夫本来在的位置那一步
}
return ;
}
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