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 MBSubmit: 634 Solved: 310
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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.
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.
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:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }