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POJ 3278 Catch That Cow BFS

1.题意:给定两个数N,K,N表示初始状态,K表示目标状态,有三种操作:+1,-1和*2,试求这个过程的最少的操作次数;

2.输入输出:依次给出N,K;

3.分析:典型的BFS求最小操作数,唯一注意的就是搜索过程中注意状态值不要超出K的范围:[0,1e5];

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <queue>
 4 using namespace std;
 5 const int maxn=1e5+10;
 6 int N,K;
 7 int vis[maxn];
 8 struct Node
 9 {
10     int X,T;
11     Node(){}
12     Node(int x,int t)
13     {
14         X=x;
15         T=t;
16     }
17 };
18 void Init()
19 {
20     for(int i=0;i<maxn;i++)
21         vis[i]=0;
22     vis[N]=1;
23 }
24 void Solve()
25 {
26     int ans=-1;
27     queue<Node> q;
28     q.push(Node(N,0));
29     while(!q.empty())
30     {
31         Node temp=q.front();
32         q.pop();
33         if(temp.X==K) 
34         {
35             ans=temp.T;
36             break;
37         }
38         if(temp.X+1<maxn&&!vis[temp.X+1])
39         {
40             vis[temp.X+1]=1;
41             q.push(Node(temp.X+1,temp.T+1));
42         }
43         if(temp.X-1>=0&&!vis[temp.X-1])
44         {
45             vis[temp.X-1]=1;
46             q.push(Node(temp.X-1,temp.T+1));
47         }
48         if(temp.X*2<maxn&&!vis[temp.X*2])
49         {
50             vis[temp.X*2]=1;
51             q.push(Node(temp.X*2,temp.T+1));
52         }
53     }
54     printf("%d\n",ans);
55 }
56 int main()
57 {
58     while(scanf("%d%d",&N,&K)!=EOF)
59     {
60         Init();
61         Solve();
62     }
63     return 0;
64 }

 

POJ 3278 Catch That Cow BFS