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

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

 

这道题和以前做的一道题很像，但是不明白那道题为什么是DP，但是似乎有人用SPFA做出来了，%用SPFA做的大神。。。

不过这道题是用SPFA，其实比较重要的就是判断2*i的边界，容易想到，如果2*i>(假设n<k;n1是n关于k对称的点)n1，那么在n1之后的点如果通过2*i

走到，那么耗费的时间一定大于直接从n走到k，所以只需要从0到n1相邻的点建立双向边，1到n1建立i*2的单向边，然后求SPFA即可。。。

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define maxn 201000
#define maxm 400000
#define ll long long
#define inf 1000000000
using namespace std;
struct edge{int go,next,w;}e[2*maxm];
int n,m,k,s,t,tot,q[maxn],d[maxn],head[maxn];
bool v[maxn];
void ins(int x,int y)
{
    e[++tot].go=y;e[tot].w=1;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=1;i<=m;++i) d[i]=inf;
    memset(v,0,sizeof(v));
    int l=0,r=1,x,y;q[1]=n;d[n]=0;
    while(l!=r)
    {
        x=q[++l];if(l==maxn)l=0;v[x]=0;
        for(int i=head[x];i;i=e[i].next)
         if(d[x]+e[i].w<d[y=e[i].go])
         {
             d[y]=d[x]+e[i].w;
             if(!v[y]){v[y]=1;q[++r]=y;if(r==maxn)r=0;}
         }
    }
}
int main(){
    scanf("%d%d",&n,&k);
    if(k<=n){printf("%d",abs(k-n));return 0;}
    m=k+abs(k-n)+1;
    for(int i=0;i<m;i++)insert(i,i+1);
    for(int i=1;i<=m/2;i++)ins(i,i*2);
    spfa();
    printf("%d",d[k]);
    return 0;
}