C - Catch That Cow
题目链接:
题目:
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?
Input
Line 1: Two space-separated integers: N and K
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input 5 17Sample Output
4 题意: 农夫约翰已被告知逃亡牛的位置,并希望立即抓住她。 他从数字线上的N点(0≤N≤100,000)开始,并且母牛在同一数字线上的点K(0≤K≤100,000)。 农夫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何一点X移动到X - 1或X + 1点 *传送:FJ可以在一分钟内从任意点X移动到2×X点。 如果母牛不知道它的追求,根本不动,那么农夫约翰需要多长时间才能找回它? 输入 第1行:两个以空格分隔的整数:N和K. 产量 第1行:Farmer John捕捉逃亡牛所需的最短时间(以分钟为单位)。 思路:广搜
#include#include #include #include using namespace std;bool book[maxn];int fan[maxn];queue qu;int bfs(int N,int K){ int tou,xia;//tou是现在位置,xia是下一步的位置 qu.push(N);//把现在位置压入队列中, fan[N]=0;//一开始步数为0 book[N]=true;//该位置已经被走过了 while(!qu.empty()) { tou=qu.front();//将现在的位置存入队列中 qu.pop();//删去队首 //cout< < 100000)//防止越界 continue; if(!book[xia]) { qu.push(xia);//把下一步的位置存入队列中 fan[xia]=fan[tou]+1; book[xia]=true;//已经走过 } } } return fan[xia];}int main(){ int N,K; while(cin>>N>>K){ memset(fan,0,sizeof(fan)); memset(book,false,sizeof(book)); while(!qu.empty()) { qu.pop(); } cout< <