POJ3278——Catch That Cow

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

USACO 2007 Open Silver

終于A掉這道題，BFS新認知。

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<vector>
#include<cmath>
#include<cstring>
#include<string> #define N 100010using 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，步數 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;
}