C – Catch That CowTime Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uSubmit Status Practice POJ 3278DescriptionFarmer 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 Output4Hint
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<iostream>#include<cstring>#include<string>#include<cmath>#include<map>#include<queue>#include<cstdio>#include<vector>#include<algorithm>#define bug printf("—–\n");using namespace std;const int maxn=100011;const int inf=210000;typedef long long ll;int d[maxn];bool vis[maxn];//从当前的X出发，可以到达X+1，X-1，，X*2，三个位置处，明显X-1不能成为负数，x+1 也不能大于K，如果大于了就会增加步数，//但是对于X*2这个条件，可以在保证X<K的情况下，计算X*2的值，而不是保证X*2的值小于K void bfs(int x,int k){queue<int> q;memset(vis,false,sizeof(vis));memset(d,0,sizeof(d));q.push(x);vis[x]=true;d[x]=0;while(!q.empty()){int x=q.front();q.pop();if(x==k)return ;if(x-1>=0&&!vis[x-1]){vis[x-1]=true;q.push(x-1);d[x-1]=d[x]+1;}if(x+1<=k&&!vis[x+1]){vis[x+1]=true;q.push(x+1);d[x+1]=d[x]+1;}if(x<=k&&x*2<=100000&&!vis[x*2]){vis[x*2]=true;q.push(x*2);d[x*2]=d[x]+1;}}}int main(){int n,m,i,j,t,k;//freopen("in.txt","r",stdin);while(~scanf("%d%d",&n,&k)){bfs(n,k);printf("%d\n",d[k]);}return 0;}
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