poj 3278 Catch That Cow
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Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 71890 Accepted: 22628
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
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.
Source
USACO 2007 Open Silver
大致题意:
给定两个整数n和k
通过 n+1或n-1 或n*2 这3种操作,使得n==k
输出最少的操作次数
/*BFS状态扩展 { 只要找到解就一定是最优的. 同样这题我们也可以用DFS跑回溯(显然要爆T) 其实迭代搜也能跑(还不会orz.) } */#include<iostream>#include<cstdio>#include<algorithm>#include<queue>#define MAXN 1000001using namespace std;int n,k,s[MAXN];bool b[MAXN];queue<int>qq;int bfs(){ qq.push(n); s[n]=0; b[n]=true; if(n>=k) return n-k; while(!qq.empty()) { int x=qq.front(),y; qq.pop(); for(int i=1;i<=3;i++) { if(i==1) y=x-1; if(i==2) y=x+1; if(i==3) y=x*2; if(y<0||y>MAXN) continue; if(!b[y]) { s[y]=s[x]+1; b[y]=true; qq.push(y); } if(y==k) return s[y]; } }}int main(){ cin>>n>>k; printf("%d",bfs()); return 0;}
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