POJ 3278 Catch That Cow （有思路有细节的bfs）

时间：2021-04-15 00:24:12

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

别看题简单，满满的坑啊！！你初始在n，先让你移动到k，有三种移动，—1，+1，*2。问你几步能到k。
最开始无脑搜，直接T了。后来发现bfs虽然是有点暴力的感觉但是还是要有思维在里面的！如果n>k，n只能一点点减到k。还有数组开大点，以防*2后越界。
代码如下：
 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <queue>
 7 using namespace std;
 8 int n,k,ans;
 9 bool vis[400010];
10 struct node
11 {
12     int pos,step;
13 };
14 bool check (int x)
15 {
16     if (vis[x])
17     return 0;
18     if (x<0||x>400010)
19     return 0;
20     else
21     return 1;
22 }
23 void bfs (node now)
24 {
25 
26     queue<node>q;
27     q.push(now);
28     while (!q.empty())
29     {
30         node temp=q.front();
31         q.pop();
32         if (check(temp.pos-1))
33         {
34             node x;
35             x.pos=temp.pos-1;
36             x.step=temp.step+1;
37             vis[x.pos]=1;
38             if (x.pos==k)
39             {
40                 ans=x.step;
41                 return ;
42             }
43             q.push(x);
44         }
45         if (check(temp.pos+1)&&temp.pos<k)
46         {
47             node x;
48             x.pos=temp.pos+1;
49             x.step=temp.step+1;
50             vis[x.pos]=1;
51             if (x.pos==k)
52             {
53                 ans=x.step;
54                 return ;
55             }
56             q.push(x);
57         }
58         if (check(temp.pos*2)&&temp.pos<k)
59         {
60             node x;
61             x.pos=temp.pos*2;
62             x.step=temp.step+1;
63             vis[x.pos]=1;
64             if (x.pos==k)
65             {
66                 ans=x.step;
67                 return ;
68             }
69             q.push(x);
70         }
71     }
72 }
73 int main()
74 {
75     //freopen("de.txt","r",stdin);
76     while (~scanf("%d%d",&n,&k))
77     {
78         memset(vis,false,sizeof vis);
79         node now;
80         now.pos=n;
81         now.step=0;
82         ans=0;
83         if (n>=k)
84         printf("%d\n",n-k);
85         else
86         {
87             bfs(now);
88             printf("%d\n",ans);
89         }
90     }
91     return 0;
92 }

 
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