Catch That Cow
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 84880 Accepted: 26651点击打开链接DescriptionFarmer 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?InputLine 1: Two space-separated integers: N and KOutputLine 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.Sample Input5 17Sample Output4HintThe 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.SourceUSACO 2007 Open Silver
题意：这道题的意思是：在一条线上有两个点n,k；其中从n点往前走（即k点走），有两种方法：
1、走一步距离1m（可以往前，可以往后），一秒
2、走2*1m（可以往前，可以往后），一秒
问：花费时间最少。
#include<iostream>#include<cstring>#include<queue>#include<cstdio>using namespace std;const int N=1000000;int map[N+10];int n,k;struct node{int x,step;};int check(int x){ if(x<0||x>=N||map[x]) return 0; return 1;}int bfs(int x){ int i; queue<node>Q; node a,next; a.x=x; a.step=0; map[x]=1; Q.push(a); while(!Q.empty()) { a=Q.front(); Q.pop(); if(a.x==k) return a.step; next=a; next.x=a.x+1; if(check(next.x)) { next.step=a.step+1; map[next.x]=1; Q.push(next); } next.x=a.x-1; if(check(next.x)) { next.step=a.step+1; map[next.x]=1; Q.push(next); } next.x=a.x*2; if(check(next.x)) { next.step=a.step+1; map[next.x]=1; Q.push(next); } } return -1;}int main(){ int ans; while(scanf("%d%d",&n,&k)!=EOF) { memset(map,0,sizeof(map)); ans=bfs(n); PRintf("%d/n",ans); } return 0;}
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