杭電 HDU 2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8999 Accepted Submission(s): 2837


Problem Description 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

HintThe 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.


睡前一水~ 好久沒做過搜索了 注意剪枝&&&&&&&&

#include #include #include #include #include #include #include #include using namespace std; struct Node { int x,step; Node(){} Node(int x,int step):x(x),step(step){} }; int n,k; bool vis[199999]; int bfs() { queueQ; Node cur,next; Q.push(Node(n,0)); while(!Q.empty()) { cur=Q.front(); Q.pop(); for(int i=0;i<3;i++) { if(i==0) next.x=cur.x+1; if(i==1) next.x=cur.x-1; if(i==2) next.x=cur.x*2; next.step=cur.step+1; if(next.x==k) return next.step; if(next.x<0||next.x>100000) continue; if(!vis[next.x]) { vis[next.x]=true; Q.push(next); } } } } int main() { while(cin>>n>>k) { memset(vis,0,sizeof(vis)); if(nk) cout<