POJ3278——Catch That Cow
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 114140 | Accepted: 35715 |
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
终于A掉这道题,BFS新认知。
#include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<vector> #include<cmath> #include<cstring> #include<string> #define N 100010 using namespace std; void in(int &x){ register char c=getchar();x=0;int f=1; while(!isdigit(c)){if(c=='-') f=-1;c=getchar();} while(isdigit(c)){x=x*10+c-'0';c=getchar();} x*=f; } int a,b,ans; struct Step{ int v,w;//位置,步数 Step(int xx,int s):v(xx),w(s){ } }; bool vis[N]; queue<Step>Q; void BFS(){ memset(vis,0,sizeof(vis)); vis[a]=1;Q.push(Step(a,0)); while(!Q.empty()){ Step s=Q.front();Q.pop(); if(s.v==b) { ans=s.w;break; } else { if(s.v-1>=0 && !vis[s.v-1]){ Q.push(Step(s.v-1,s.w+1)); vis[s.v-1]=1; }if(s.v+1<=N &&!vis[s.v+1]){ Q.push(Step(s.v+1,s.w+1)); vis[s.v+1]=1; }if(s.v*2<=N && !vis[s.v*2]){ Q.push(Step(s.v*2,s.w+1)); vis[s.v*2]=1; } } } } int main() { in(a);in(b); BFS(); printf("%d\n",ans); return 0; }
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