poj3278_kuagnbin带你飞专题一
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 88247 | Accepted: 27640 |
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
做对一个题是如此的不容易。。。。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; int que[200001]; int start=0,endd=0; int time[200001]={0}; int vis[200001]={0}; int n,k; int bfs(int n,int k){ memset(que,0,sizeof(que)); memset(time,0,sizeof(time)); memset(vis,0,sizeof(vis)); start=0; endd=0; que[endd++]=n; vis[n]=1; while(start<endd){ int t=que[start]; start++; for(int i=0;i<3;i++){ int tt=t; if(i==0){ tt+=1; }else if(i==1){ tt-=1; }else if(i==2){ tt*=2; } if(tt>100000||tt<0){//注意一定要排除>100000的情况，不然会re，也不是>k continue; } if(!vis[tt]){ vis[tt]=1; que[endd]=tt; time[tt]=time[t]+1; if(tt==k){ return time[tt]; } endd++; } } } } int main() { int ans; while(~scanf("%d %d",&n,&k)){ if(n<k){ ans=bfs(n,k); }else{ ans=n-k; } printf("%d\n",ans); } return 0; }
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