Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 97240 | Accepted: 30519 |
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.
思路：bfs入门题，用到一些剪枝，之前没怎么用bfs不是很熟练，看了些其他大佬的代码，每个点都有三种情况，将三种情况入队，然后标记这个点已经访问，最后最先达到k点的必为最短。直接输出就行了
实现代码：
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<list>
using namespace std;
#define me0(x) memset(x,0,sizeof(x))
#define pb(x) push_back(x)
#define ll long long
const int Mod = 1e9+7;
const int inf = 1e9;
const int Max = 2e5+10;
vector<int>vt[Max];
queue<int>q;
int dx[] = {-1, 1, 0, 0};
int dy[] = { 0, 0, -1, 1};
//void exgcd(ll a,ll b,ll& d,ll& x,ll& y){if(!b){d=a;x=1;y=0;}else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}}
//ll inv(ll a,ll n){ll d, x, y;exgcd(a,n,d,x,y);return (x+n)%n;} ???
//int gcd(int a,int b) { return (b>0)?gcd(b,a%b):a; } ??С???
//int lcm(int a, int b) { return a*b/gcd(a, b); } ??С????
int cnt[Max];
bool vis[Max];
void bfs(int l,int r)
{
q.push(l);
vis[l] = 1;
cnt[l] = 0;
while(!q.empty()){
int x = q.front();
q.pop();
if(x==r){
cout<<cnt[r]<<endl;
break;
}
if(x-1>=0&&!vis[x-1]){
vis[x-1] = 1;
q.push(x-1);
cnt[x-1] = cnt[x] + 1;
}
if(x<=r&&!vis[x+1]){
vis[x+1] = 1;
q.push(x+1);
cnt[x+1] = cnt[x] + 1;
}
if(x<=r&&!vis[x*2]){
vis[x*2] = 1;
q.push(x*2);
cnt[x*2] = cnt[x] + 1;
}
}
}
int main()
{
int n,m;
cin>>n>>m;
me0(vis);
bfs(n,m);
return 0;
}