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

题解:
构图完了SPFA。。。
代码:
 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 200000+1000
14 #define maxm 400000
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 using namespace std;
19 inline int read()
20 {
21     int x=0,f=1;char ch=getchar();
22     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
23     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
24     return x*f;
25 }
26 struct edge{int go,next,w;}e[2*maxm];
27 int n,s,t,tot,q[maxn],d[maxn],head[maxn];
28 bool v[maxn];
29 void ins(int x,int y)
30 {
31     e[++tot].go=y;e[tot].next=head[x];head[x]=tot;
32 }
33 void insert(int x,int y)
34 {
35     ins(x,y);ins(y,x);
36 }
37 void spfa()
38 {
39     for(int i=0;i<=n;++i) d[i]=inf;
40     memset(v,0,sizeof(v));
41     int l=0,r=1,x,y;q[1]=s;d[s]=0;
42     while(l!=r)
43     {
44         x=q[++l];if(l==maxn)l=0;v[x]=0;
45         for(int i=head[x];i;i=e[i].next)
46          if(d[x]+1<d[y=e[i].go])
47          {
48              d[y]=d[x]+1;
49              if(!v[y]){v[y]=1;q[++r]=y;if(r==maxn)r=0;}
50          }
51     }
52     
53 }
54 int main()
55 {
56     freopen("input.txt","r",stdin);
57     freopen("output.txt","w",stdout);
58     s=read();t=read();
59     if(s>=t){printf("%d\n",abs(t-s));return 0;}
60     else n=t+abs(t-s)+1;
61     for(int i=0;i<n;i++)insert(i,i+1);
62     for(int i=1;i<=n/2;i++)ins(i,i<<1);
63     spfa();
64     printf("%d\n",d[t]);
65     return 0;
66 }
View Code

 

posted @ 2014-08-28 13:44  ZYF-ZYF  Views(174)  Comments(0Edit  收藏  举报