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?
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 17Sample Output
4Hint
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.
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
#include <vector>
using namespace std;
typedef long long ll;
typedef struct node {
int x, step;
} node;
const int maxn = 1e5 + 10;
int t, n, m, ans, cnt, sx, sy, sz, ex, ey, ez;
char mp[35][35][35];
int vis[maxn];
bool OutLimit(int x) {
if(x < 0 || x > 100000) {
return 1;
}
return 0;
}
int BFS() {
queue<node> Q;
node a, b;
a.x = n;
a.step = 0;
vis[a.x] = 1;
Q.push(a);
while(!Q.empty()) {
a = Q.front();
Q.pop();
if(a.x == m) {
return a.step;
}
for(int i = 0; i < 3; i++) {
if(i == 0) {
b.x = a.x - 1;
} else if(i == 2) {
b.x = a.x + 1;
} else {
b.x = 2 * a.x;
}
b.step = a.step + 1;
if(OutLimit(b.x)) {
continue;
}
if(!vis[b.x]) {
vis[b.x] = 1;
Q.push(b);
}
}
}
return 0;
}
int main() {
ios::sync_with_stdio(false);
cin >> n >> m;
int ans = BFS();
cout << ans << endl;
return 0;
}