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poj3278(Catch That Cow)

题目地址:Catch That Cow

 

题目大意:

    一位农夫追赶一头牛,题目给出农夫和牛的坐标分别为N,K。农夫可以通过坐标的加一或减一也可以坐标乘以2。问你最少多少步到达牛的坐标。

 

解题思路;

   简单BFS。

 

代码:

 1 #include <algorithm> 2 #include <iostream> 3 #include <sstream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <string> 8 #include <bitset> 9 #include <vector>10 #include <queue>11 #include <stack>12 #include <cmath>13 #include <list>14 //#include <map>15 #include <set>16 using namespace std;17 /***************************************/18 #define ll long long19 #define int64 __int6420 #define PI 3.141592721 /***************************************/22 const int INF = 0x7f7f7f7f;23 const double eps = 1e-8;24 const double PIE=acos(-1.0);25 const int d1x[]= {0,-1,0,1};26 const int d1y[]= {-1,0,1,0};27 const int d2x[]= {0,-1,0,1};28 const int d2y[]= {1,0,-1,0};29 const int fx[]= {-1,-1,-1,0,0,1,1,1};30 const int fy[]= {-1,0,1,-1,1,-1,0,1};31 const int dirx[]= {-1,1,-2,2,-2,2,-1,1};32 const int diry[]= {-2,-2,-1,-1,1,1,2,2};33 /*vector <int>map[N];map[a].push_back(b);int len=map[v].size();*/34 /***************************************/35 void openfile()36 {37     freopen("data.in","rb",stdin);38     freopen("data.out","wb",stdout);39 }40 priority_queue<int> qi1;41 priority_queue<int, vector<int>, greater<int> >qi2;42 /**********************华丽丽的分割线,以上为模板部分*****************/43 int vis[200009],cnt[200009];44 int BFS(int n,int k)45 {46     queue<int >Q;47     Q.push(n);48     int v;49     while(!Q.empty())50     {51         v=Q.front();52         Q.pop();53         if (v==k)54             return cnt[k];55         if (v<100005&&v>-2)56         {57             if (!vis[v-1])58             {59                 Q.push(v-1);60                 if (!cnt[v-1])61                     cnt[v-1]=cnt[v]+1;62                 vis[v-1]=1;63             }64             if (!vis[v+1])65             {66                 Q.push(v+1);67                 if (!cnt[v+1])68                     cnt[v+1]=cnt[v]+1;69                 vis[v+1]=1;70             }71             if (!vis[v*2])72             {73                 Q.push(v*2);74                 if (!cnt[v*2])75                     cnt[v*2]=cnt[v]+1;76                 vis[v*2]=1;77             }78         }79     }80     return 0;81 }82 int main()83 {84 85     int n,k,sum;86     while(scanf("%d%d",&n,&k)!=EOF)87     {88         memset(vis,0,sizeof(vis));89         memset(cnt,0,sizeof(cnt));90         sum=BFS(n,k);91         printf("%d\n",sum);92     }93     return 0;94 }
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