Catch That Cow
TimeLimit: 2000ms MemoryLimit:65536KB
64-bit integer IO format:%lld
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.
SampleInput
5 17
SampleOutput
4
[思路]：这题就是个bfs，很水的bfs
因为他是一维的，
所以只要考虑+1，-1，*2这三种情况
还有如果n>k那么就只能倒退
输出n-k就ojbk了
贴上代码：
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #define MAXN 200005 using namespace std; int vids[MAXN]; int step[MAXN]; queue<int>q; int bfs(int n,int k) { int head,ends; q.push(n); while(!q.empty()) { head=q.front(); q.pop(); for(int i=0; i<3; i++) { if(i==0) ends=head-1; if(i==1) ends=head+1; if(i==2) ends=head*2; if(ends<0||ends>MAXN)continue; if(vids[ends]==0) { q.push(ends); vids[ends]=1; step[ends]=step[head]+1; } if(ends==k) { return step[ends]; } } } } int main() { int n,k; int sum; while(~scanf("%d%d",&n,&k)) { memset(step,0,sizeof(step)); memset(vids,0,sizeof(vids)); if(n>=k)printf("%d ",n-k); else { sum=bfs(n,k); printf("%d ",sum); while(!q.empty()) q.pop(); } } return 0; }