Poj 3278: Catch that cow (simple one-dimensional wide search)

Catch that cow

Time limit:2000 ms		Memory limit:65536 K
Total submissions:45648		Accepted:14310

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.

Source

Usaco 2007 Open silver Simple one-dimensional wide search. Question: You are at the N position of a one-dimensional coordinate, and the ox is at the K position. You need to capture the ox from N to K. You can have three steps: n + 1, n-1, or N * 2. Please catch the shortest step of the cow. Ideas: The idea of simple and extensive search. There are three jump methods for status redirection. Place the new status in the queue and extract the most advanced status in the queue until the K position is found. Code:

1 # include <iostream> 2 # include <stdio. h> 3 # include <string. h> 4 # include <queue> 5 using namespace STD; 6 7 bool ISW [100010]; 8 9 struct node {10 int X; 11 int s; 12 }; 13 14 bool judge (int x) 15 {16 if (x <0 | x> 100000) 17 return true; 18 if (ISW [x]) 19 Return true; 20 return false; 21} 22 23 int BFS (int sta, int end) 24 {25 queue <node> q; 26 node cur, next; 27 cur. X = sta; 28 cur. S = 0; 29 ISW [cur. x] = true; 30 Q. Push (cur); 31 While (! Q. empty () {32 cur = Q. front (); 33 Q. pop (); 34 if (cur. X = END) 35 return cur. s; 36 // start with 37 int NX; 38 Nx = cur. X + 1; 39 if (! Judge (nx) {40 next. X = NX; 41 next. S = cur. S + 1; 42 ISW [next. x] = true; 43 Q. push (next); 44} 45 Nx = cur. x-1; 46 If (! Judge (nx) {47 next. X = NX; 48 next. S = cur. S + 1; 49 ISW [next. x] = true; 50 Q. push (next); 51} 52 // jump forward 53 Nx = cur. x * 2; 54 if (! Judge (nx) {55 next. X = NX; 56 next. S = cur. S + 1; 57 ISW [next. x] = true; 58 Q. push (next); 59} 60} 61 Return 0; 62} 63 64 65 int main () 66 {67 int N, K; 68 while (scanf ("% d", & N, & K )! = EOF) {69 memset (ISW, 0, sizeof (ISW); 70 int step = BFS (n, k); 71 printf ("% d \ n ", step); 72} 73 return 0; 74}

 

Freecode: www.cnblogs.com/yym2013