[BZOJ] 1646: [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: 1266  Solved: 599
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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.

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

 

Analysis

BFS！

 

Code

#include<cstdio>
#include<iostream>
#include<queue>
#define maxn 1010101
using namespace std;

bool judge(int x){
    return x < maxn && x >= 0;
}

struct node{
    int pos,step;
};

bool book[maxn];
int n,k;

void bfs(){
    queue<node> Q;
    Q.push((node){n,0});
    book[n] = true;
    
    while(!Q.empty()){
        node now = Q.front();
        Q.pop();
        
        if(now.pos == k){
            printf("%d",now.step);
            return;
        }
        
        if(judge(now.pos+1) && !book[now.pos+1]){
            book[now.pos+1] = true;
            Q.push((node){now.pos+1,now.step+1});
        }
        
        if(judge(now.pos-1) && !book[now.pos-1]){
            book[now.pos-1] = true;
            Q.push((node){now.pos-1,now.step+1});
        }
        
        if(judge(now.pos*2) && !book[now.pos*2]){
            book[now.pos*2] = true;
            Q.push((node){now.pos*2,now.step+1});
        }
    }
}

int main(){
    scanf("%d%d",&n,&k);
    
    book[n] = true;
    bfs();
    
    return 0;
}