# [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

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.

Silver

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;
}

posted @ 2017-09-04 17:54  Leviaton  阅读(152)  评论(0编辑  收藏  举报