Time Limit: 2000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
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
Output
Sample Input
5 17
Sample Output
4
Hint
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;
}