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

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

Time Limit: 5 Sec Memory Limit: 128 MB

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?

农夫约翰得知了一头逃亡的母牛的位置，他想立即抓住她。他从数字线上的一个点n（0<=n<=100000）开始，奶牛在同一条数字线上的一个点k（0<=k<=100000）。农民约翰有两种交通方式：步行和传送。步行：FJ可以在一分钟内从任意点X移动到点X-1或X+1传送：FJ可以在一分钟内从任意点X移动到点2*X。如果牛没有意识到它的追求，一点也不动，农夫约翰需要多长时间才能找回它？

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

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.

code:

#include<iostream> 
#include<cstdio> 
#include<cmath> 
#include<cstring> 
#include<cstdlib> 
#include<algorithm> 
#include<iomanip> 
#include<map> 
#include<set> 
#include<vector> 
#include<queue> 
using namespace std; 
inline int read() 
{ 
    int x=0,f=1;char ch=getchar(); 
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 
    while(ch<='9'&&ch>='0'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();} 
    return x*f; 
} 
const int N=1e5+100; 
bool mark[N<<1],flag=1; 
int n,m,q[N<<2],dis[N<<1],ans; 
int main() 
{ 
    n=read();m=read();  if(n>m){printf("%d\n",n-m);return 0;} 
    int hd=0,tl=2; 
    q[++hd]=n;mark[n]=0;q[tl++]=m;mark[m]=1; 
    dis[1]=dis[2]=0; 
    while(hd<tl) 
    { 
        int t=q[hd];hd++; 
        if(!mark[t]) 
        { 
            if(t-1>=0&&!dis[t-1]&&!mark[t-1]) 
            {dis[t-1]=dis[t]+1;mark[t-1]=0;q[tl++]=t-1;} 
            else if(t-1>=0&&mark[t-1]) 
            {ans=dis[t]+dis[t-1]+1;break;} 
            if(t+1<=n+m&&!dis[t+1]&&!mark[t+1]) 
            {dis[t+1]=dis[t]+1;mark[t+1]=0;q[tl++]=t+1;} 
            else if(t+1<=n+m&&mark[t+1]) 
            {ans=dis[t]+dis[t+1]+1;break;} 
            if(t<<1<=n+m&&!dis[t<<1]&&!mark[t<<1]) 
            {dis[t<<1]=dis[t]+1;mark[t<<1]=0;q[tl++]=t<<1;} 
            else if(t<<1<=n+m&&mark[t<<1]) 
            {ans=dis[t]+dis[t<<1]+1;break;} 
        } 
        else
        { 
            if(t-1>=0&&!dis[t-1]) 
            {dis[t-1]=dis[t]+1;mark[t-1]=1;q[tl++]=t-1;} 
            else if(t-1>=0&&!mark[t-1]) 
            {ans=dis[t]+dis[t-1]+1;break;} 
            if(t+1<=n+m&&!dis[t+1]) 
            {dis[t+1]=dis[t]+1;mark[t+1]=1;q[tl++]=t+1;} 
            else if(t+1<=n+m&&!mark[t+1]) 
            {ans=dis[t]+dis[t+1]+1;break;} 
            if(!(t&1)&&flag) 
            { 
                if(t==0)flag=0; 
                if(!dis[t>>1]) 
                {dis[t>>1]=dis[t]+1;mark[t>>1]=1;q[tl++]=t>>1;} 
                else if(!mark[t>>1]) 
                {ans=dis[t]+dis[t>>1]+1;break;} 
            } 
        } 
    } 
    printf("%d\n",ans); 
}