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

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

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 ;
 }