poj 3278 Catch That Cow 广搜

93 阅读 0 评论 62 点赞

我是靠谱客的博主唠叨电灯胆，这篇文章主要介绍poj 3278 Catch That Cow 广搜，现在分享给大家，希望可以做个参考。

hdu 2717 Catch That Cow,题目链接

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11466 Accepted Submission(s): 3551

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.

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

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.

Source

USACO 2007 Open Silver

Recommend

teddy | We have carefully selected several similar problems for you: 2102 1372 1240 1072 1180

题意：给你两个位置k和n，求出从k到n的最小步数。
从k有三种方式变化方式： k+1，k-1，k*2。

队列广搜，一个vis数组查看是否访问过，一个step数组记录步数。

复制代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
#include<stdio.h>
#include<queue>
#include<algorithm>
#include<string>
#include<iostream>
#include<string.h>
#include<math.h>
#define maxn 100005
using namespace std;
int vis[maxn];//标记是否访问过;
int step[maxn];//记录步数;
queue <int > q ;
int serch(int k,int n)
{
int flag=0,a,mark=0;
memset(vis,0,sizeof(vis));//数组清零;
memset(step,0,sizeof(step));//数组清零;
q.push(k);
//将k入队;
vis[k]=1;
//标记为1;
while(!q.empty())
//当数组非空时;
{
a=q.front();
//将队列首位赋值给a;
q.pop();
//删除队列首位;
for(int i=1; i<=3; i++)
{
if(i==1)
flag=a+1;
else if(i==2)
flag=a-1;
else
flag=a*2;
if(flag>100000||flag<0)//越界判断;
continue;
if(!vis[flag])//如果标记为0;
{
step[flag]=step[a]+1;//当前步数等于上次步数+1;
vis[flag]=1;//标记为1;
q.push(flag); //入队;
if(flag==n)//如果当前值为n;
return step[flag];//返回步数;
}
}
}
return 0;
//如果是提交C++的话，要不要返回值都可以;
//但如果是G++的话，必须要一个return 0;
//这个是编译器的问题，具体也不是很清楚;
}
int main()
{
int n,k;
//k追n;
while(~scanf("%d%d",&k,&n))//在这里我写成了k追n;
{
if(k>n) //如果k>n，只能通过不断减一得到，也就是k-n步;
printf("%dn",k-n);
else
printf("%dn",serch(k,n));
}
return 0;
}