BZOJ1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 634 Solved: 310
[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

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

Silver

题解：

构图完了SPFA。。。

代码：

 #include<cstdio>

 #include<cstdlib>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<iostream>

 #include<vector>

 #include<map>

 #include<set>

 #include<queue>

 #include<string>

 #define inf 1000000000

 #define maxn 200000+1000

 #define maxm 400000

 #define eps 1e-10

 #define ll long long

 #define pa pair<int,int>

 using namespace std;

 inline int read()

 {

     int x=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }

 struct edge{int go,next,w;}e[*maxm];

 int n,s,t,tot,q[maxn],d[maxn],head[maxn];

 bool v[maxn];

 void ins(int x,int y)

 {

     e[++tot].go=y;e[tot].next=head[x];head[x]=tot;

 }

 void insert(int x,int y)

 {

     ins(x,y);ins(y,x);

 }

 void spfa()

 {

     for(int i=;i<=n;++i) d[i]=inf;

     memset(v,,sizeof(v));

     int l=,r=,x,y;q[]=s;d[s]=;

     while(l!=r)

     {

         x=q[++l];if(l==maxn)l=;v[x]=;

         for(int i=head[x];i;i=e[i].next)

          if(d[x]+<d[y=e[i].go])

          {

              d[y]=d[x]+;

              if(!v[y]){v[y]=;q[++r]=y;if(r==maxn)r=;}

          }

     }

 }

 int main()

 {

     freopen("input.txt","r",stdin);

     freopen("output.txt","w",stdout);

     s=read();t=read();

     if(s>=t){printf("%d\n",abs(t-s));return ;}

     else n=t+abs(t-s)+;

     for(int i=;i<n;i++)insert(i,i+);

     for(int i=;i<=n/;i++)ins(i,i<<);

     spfa();

     printf("%d\n",d[t]);

     return ;

 }