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[题解]|[题解] Catch That Cow 抓住那只牛 C++


Catch That Cow 抓住那只牛

  • 题目
    • Description
    • Input
    • Output
      • Sample Input
      • Sample Output
  • 思路
  • 代码

题目 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x处走到x+1或x-1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
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.
思路 【[题解]|[题解] Catch That Cow 抓住那只牛 C++】典型bfs板子题,且是在一维数组上进行，每次for0~1扩展前进1布和后退1布，再扩展2x步，判断是否出界标记
要是数组刚好100000，就算你判断了边界也会RE,再多开一半就没事了
代码
#include using namespace std; int n,k; int dir[2]={1,-1}; struct node{ int x; int d; node(int xx,int dd){ x=xx; d=dd; } }; bool in(int x){ return x>=0 && x<=150000; } bool vis[150005]; int bfs(int sx){ queue q; if(sx==k){ return 0; } q.push(node(sx,0)); while(!q.empty()){ node now=q.front(); //q.pop(); for(int i=0; i<2; i++){ int tx=now.x+dir[i]; if(in(tx) && !vis[tx]){ vis[tx]=1; if(tx==k){ return now.d+1; } q.push(node(tx,now.d+1)); } } int tx=2*now.x; if(in(tx) && !vis[tx]){ vis[tx]=1; if(tx==k){ return now.d+1; } q.push(node(tx,now.d+1)); } } } int main(){ while(1); string s; cin>>n>>k; getline(cin,s); cout<

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