POJ_3278 Catch That Cow 解题报告
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 28177 | Accepted: 8677 |
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 17
Sample Output
4
Hint
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.
题目连接： http://poj.org/problem?id=3278
算法类型：BFS
解题思路：BFS搜索，标记已经访问的进行剪枝。
算法实现：
#include
#include
#include
int vis[100000]; //记录是否访问过；
int dis[100000]; //记录步数；
int q[100000]; //队列；
int n,k;
int bfs(int n) //广搜函数；
{
int rear=0 ,front=0,u; //队列头和尾都为0；
vis[n]==1;
q[rear++]=n; //入队列；
while(front0) //下面三个IF 是广搜的三个方向；
{
dis[u*2]=dis[u]+1; //记录达到这点走过的步数；
q[rear++]=u*2; //入队列；
vis[u*2]=1;
}
if(u+1<=100000 && vis[u+1]==0 && u+1>0)
{
dis[u+1]=dis[u]+1;
q[rear++]=u+1;
vis[u+1]=1;
}
if(u-1<=100000 && u-1>0 && vis[u-1]==0)
{
dis[u-1]=dis[u]+1;
q[rear++]=u-1;
vis[u-1]=1;
}
}
return 1;
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(vis,0,sizeof(vis)); //置0函数；
memset(q,0,sizeof(q));
memset(dis,0,sizeof(dis));
if(k<=n) //判断是人在前，还是牛在前；
{
printf("%d\n",n-k); // 输出步数；
}
else
{
bfs(n); //调用宽搜函数；
printf("%d\n",dis[k]); // 输出步数；
}
}
return 0;
}