POJ 3278 Catch That Cow (BFS), pojbfs

POJ 3278 Catch That Cow (BFS), pojbfs

Question URL:Http://poj.org/problem? Id = 3278

Question:

Catch That Cow

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:92360		Accepted:28999

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

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.

Ideas:

A simple BFS question. There are only three operations in total. Each operation takes the first element of the team and performs three operations respectively. Then, the operation result is judged to screen out negative numbers and more than 100,000 (boundary value, in other cases, join the team.

Code:

1 # include <cstdio> 2 # include <queue> 3 # include <algorithm> 4 using namespace std; 5 int visited [100005]; 6 int n, k; 7 queue <int> q; 8 void bfs () {9 q. push (n); 10 while (! Q. empty () {11 int x = q. front (); q. pop (); 12 if (x = k) {13 printf ("% d \ n", visited [x]-1); // the initial number of times is 1, result minus 114 return; 15} 16 if (x-1> = 0 &&! Visited [x-1]) {// negative number does not consider 17 q. push (x-1); 18 visited [x-1] = visited [x] + 1; 19} 20 if (x + 1 <= 100000 &&! Visited [x + 1]) {// If the boundary value is exceeded, it will take up unnecessary search time 21 q. push (x + 1); 22 visited [x + 1] = visited [x] + 1; 23} 24 if (x * 2 <= 100000 &&! Visited [x * 2]) {25 q. push (x * 2); 26 visited [x * 2] = visited [x] + 1; 27} 28} 29} 30 int main () {31 scanf ("% d", & n, & k); 32 visited [n] = 1; // assign a value to the initial position, prevent 33 bfs (); 34 return 0; 35} from the second search}