1646: [Usaco2007 Open]Catch That Cow 抓住那只牛
Time Limit: 5 Sec
Memory Limit: 64 MB
Submit: 732
Solved: 345
[ Submit][ Status]
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?
农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．
他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有
两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．
那么，约翰需要多少时间抓住那只牛呢？
Input
* Line 1: Two space-separated integers: N and K
仅有两个整数N和K.
Output
* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
最短的时间．
Sample Input
5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.
Sample Output
4
OUTPUT DETAILS:
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.
HINT
Source
用spfa优化dp转移。
把起点入队，每次都搜2*x,x-l,x+l，能优化的就入队，直到队列为空。
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int inq[200005],f[200005],n,k;
int main()
{
scanf("%d%d",&n,&k);
memset(f,0x3f3f3f3f,sizeof(f));
memset(inq,0,sizeof(inq));
f[n]=0;
queue<int> q;
q.push(n);
inq[n]=1;
while (!q.empty())
{
int x=q.front();
q.pop();
inq[x]=0;
if (2*x<=200000&&f[x]+1<f[2*x])
{
f[2*x]=f[x]+1;
if (!inq[2*x]) q.push(2*x),inq[2*x]=1;
}
for (int i=-1;i<=1;i++)
if (i)
{
if (x+i>=0&&x+i<=200000&&f[x+i]>f[x]+1)
{
f[x+i]=f[x]+1;
if (!inq[x+i]) q.push(x+i),inq[x+i]=1;
}
}
}
cout<<f[k]<<endl;
return 0;
}
