Catch that cow (breadth-first search _ BFS)

Catch that cow

Time limit:2000 ms		Memory limit:65536 K
Total submissions:48036		Accepted:15057

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(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 pointXTo the pointsX-1 orX+ 1 in a single minute
* Teleporting: FJ can move from any pointXTo the point 2 ×XIn 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: NAnd 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

Question: enter two numbers N and K. Calculate the minimum number of steps from N to K. You can move forward 1 to backward 1 or the current position * 2;

# Include <stdio. h> # include <string. h> # include <stdlib. h ># include <algorithm> using namespace STD; struct node {int X; // int ans at the current position; // number of steps taken} Q [1000010]; int vis [1000010]; // indicates whether the variable is accessed; int JX [] = {-}; // returns 1 or advances 1; struct node T, F; int N, K; void BFS () {int I; int S = 0, E = 0; // the pointer simulates the queue. Add e ++ to the queue and propose several s ++ memset (VIS, 0, sizeof (VIS); T. X = N; // The current initial position vis [T. x] = 1; // marked as 1 indicates that you have accessed it; T. ans = 0; // the initial number of steps is 0; Q [E ++] = T; // Add the current number of steps to the queue while (S <E) // when the queue is not empty {T = Q [s ++]; // propose if (T. X = k) // if the number is exactly equal to the number of direct output steps at the target position {printf ("% d \ n", T. ans); break;} for (I = 0; I <3; I ++) // I = 0 step back, I = 1 step forward, I = 2 position * 2 at this time; {if (I = 2) {f. X = T. x * 2;} else {f. X = T. X + JX [I];} If (F. x> = 0 & F. x <= 100000 &&! Vis [F. x]) {f. ans = T. ans + 1; Q [E ++] = f; vis [F. x] = 1 ;}}} int main () {While (~ Scanf ("% d", & N, & K) {BFS ();} return 0 ;}

Catch that cow (breadth-first search _ BFS)