C - Catch That Cow POJ - 3278

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.
 1 #include<set>
 2 #include<map>
 3 #include<cmath>
 4 #include<queue>
 5 #include<string>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<sstream>
 9 #include<cstdlib>
10 #include<cstring>
11 #include<sstream>
12 #include<iostream>
13 #include<algorithm>
14 
15 using namespace std;
16 const int maxn = 100500;
17 const int INF = 0x3f3f3f3f;
18 const int mod = 7;
19 typedef long long ll;
20 #define PI 3.1415927
21 
22 int n, k;
23 bool vis[maxn];
24 int ans;
25 struct Node
26 {
27     int num;
28     int step;
29     Node(int _num, int _s){num=_num;step=_s;}
30 };
31 void bfs()
32 {
33     queue<Node> que;
34     que.push(Node(n, 0));
35     vis[n] = true;
36     while(que.size())
37     {
38         Node nn = que.front(); que.pop();
39         int nnum = nn.num+1;
40         if(0<=nnum&&nnum<maxn&&!vis[nnum])
41         {
42             if(nnum == k)
43             {
44                 ans = nn.step+1;
45                 break;
46             }
47             que.push(Node(nnum, nn.step+1));
48             vis[nnum] = true;
49         }
50         nnum = nn.num-1;
51         if(0<=nnum&&nnum<maxn&&!vis[nnum])
52         {
53             if(nnum == k)
54             {
55                 ans = nn.step+1;
56                 break;
57             }
58             que.push(Node(nnum, nn.step+1));
59             vis[nnum] = true;
60         }
61         nnum = nn.num*2;
62         if(0<=nnum&&nnum<maxn&&!vis[nnum])
63         {
64             if(nnum == k)
65             {
66                 ans = nn.step+1;
67                 break;
68             }
69             que.push(Node(nnum, nn.step+1));
70             vis[nnum] = true;
71         }
72     }
73 }
74 int main()
75 {
76     scanf("%d%d", &n, &k);
77     if(n<k)
78     {
79         bfs();
80         printf("%d\n", ans);
81     }
82     else
83     {
84         printf("%d", n-k);
85     }
86     return 0;
87 }
AC代码

 

posted @ 2018-07-31 15:13  focus5679  阅读(95)  评论(0)  编辑  收藏  举报