POJ 3278：Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

Submit Status

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.

发现广度优先搜索适合什么题呢？就是那种在每个点都给你几个选择，然后问你最短路径的问题，对，就是这样，这样的题目最适合广度优先搜索。

这次的这个就是，每次Farmer有三个选择，然后求最短到达目的地的行程。简直是广搜的模板题。

代码：

#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <string>#include <queue>#include <cstring>#pragma warning(disable:4996)using namespace std;int color[1000005];int dis[1000005];queue<int> q;int main(){//freopen("i.txt","r",stdin);//freopen("o.txt","w",stdout);int N,K;cin>>N>>K;if(N==K){cout<<0<<endl;}else{memset(color,0,sizeof(color));memset(dis,0,sizeof(dis));q.push(N);while(!q.empty()){N=q.front();q.pop();if(N-1==K || N+1==K || 2*N==K){cout<<dis[N]+1<<endl;break;}if(color[N-1]==0 && N-1>=0){color[N-1]=1;dis[N-1]=dis[N]+1;q.push(N-1);}if(color[N+1]==0){color[N+1]=1;dis[N+1]=dis[N]+1;q.push(N+1);}if(color[2*N]==0 && 2*N<=100000){color[2*N]=1;dis[2*N]=dis[N]+1;q.push(2*N);}}}return 0;}