POJ 3278:Catch That Cow 抓住那頭牛
- 總時間限制:
- 2000ms
- 內存限制:
- 65536kB
- 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?
- 輸入
- Line 1: Two space-separated integers: N and K
- 輸出
- Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
- 樣例輸入
-
5 17
- 樣例輸出
-
4
- 提示
- 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.
/* 廣度優先搜索,重要的是剪枝和標記訪問過的路徑。 所謂的剪枝,也可以理解爲生成每一個子節點的條件;所謂條件限制越嚴謹,剪枝效果越好。 */ #include <iostream> #include <cstring> #include <queue> using namespace std; const int MAX=100020; // 這個MAX的設定並不太嚴謹 int visited[MAX]={0}; int main() { int n,k; while (cin>>n>>k){ if( n > k ){ cout<<n-k<<endl; continue; } memset(visited,0,sizeof(visited)); queue<int> q; q.push(n); int t=0; while( !q.empty() ){ t = q.front(); q.pop(); if(t == k){ break; } if( t < k && !visited[t+1] ){ // 當t>k時,不執行+1 action q.push(t+1); visited[t+1] = visited[t] + 1; } if( t > 0 && !visited[t-1] ){ // q.push(t-1); visited[t-1] = visited[t] + 1; } if( 2*t < MAX && !visited[2*t] ){ //此處不太嚴謹,如上討論 q.push(2*t); visited[2*t] = visited[t] +1; } } cout<<visited[t]<<endl; } return 0; }