【題目描述】

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.
【題目分析】
剛開始一直想用DFS做，還加上了記憶化，可是還是一直超時，對于這種求最少值 的問題，還是BFS比較有優勢，因為隊列本身自帶最少屬性（步數少加入的早，就總在前面），而且BFS沒怎么優化就過了，說明數據還是比較友好的。
代碼

#include<cstdio> #include<queue> #include<cstring> #include<climits>using namespace std;int n,k,t,m,ans,p; const int MAXN=100005; int a[MAXN];void BFS(int x) {queue<int> q;q.push(x);a[x]=0;while(!q.empty()){t=q.front(); q.pop();if(t==k){if(a[t]<ans)ans=a[t];return;}for(int i=0;i<3;i++){if(i==0) m=t-1;else if(i==1) m=t+1;else if(i==2) m=t*2;if(m<0) continue;if(m>k){p=a[t]+1+m-k;if(p<ans) ans=p;continue;}if(a[m]!=-1) continue;a[m]=a[t]+1;q.push(m);}} }int main() {scanf("%d%d",&n,&k);memset(a,-1,sizeof(a));if(n>=k){printf("%d",n-k);}else{ans=INT_MAX;BFS(n);printf("%d",ans);} }

總結

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