Catch That Cow
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 84880 Accepted: 26651點擊打開鏈接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
題意：這道題的意思是：在一條線上有兩個點n,k；其中從n點往前走（即k點走），有兩種方法：
1、走一步距離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
問：花費時間最少。
#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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