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
Output
4
Hint poj3278 有链接提示的题目请先去链接处提交程序，AC后提交到SDUTOJ中，以便查询存档。 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<stdio.h> #include<stdlib.h> #include<string.h> #include<iostream> #include<algorithm> #include<queue> using namespace std; const int N = 200100; int n, k; struct node { int x, step; }; int vis[N]; void BFS(node w1) { int X,STEP; queue<node>Q; Q.push(w1); while(!Q.empty()) { node w2= Q.front(); Q.pop(); X=w2.x; STEP=w2.step; if(X == k) { printf("%d\n",STEP); return ; } if(X >= 1 && vis[X - 1]==0)//向左走 { node w3; vis[X-1]=1; w3.x=X-1; w3.step=STEP + 1; Q.push(w3); } if(X <= k && vis[X + 1]==0) { node w3; vis[X+1] = 1; w3.x = X + 1; w3.step = STEP + 1; Q.push(w3); } if(X <= k && vis[X * 2]==0) { node w3; vis[X * 2] = 1; w3.x = 2 * X; w3.step = STEP + 1; Q.push(w3); } } } int main() { while(scanf("%d%d",&n,&k)!=EOF) { memset(vis,0,sizeof(vis)); vis[n] = 1; node w1; w1.x = n; w1.step = 0; BFS(w1); } return 0; }