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?
也是典型的BFS的题目,我第一次设的数组最大是100000*10,然后过了,第二次100000+5,居然也过了。。。
这个题就是从出发点开始bfs,有三条路可以走,一知道发现牛为止。
代码如下:
#include<iostream> #include<cstring> #include<queue> using namespace std; int N,K; int rem[100005]; void bfs() { queue <int> que; int t; que.push(N); rem[N]=0; while(!que.empty()) { t=que.front(); que.pop(); if(t==K) return; if(t*2<=100000&&rem[t*2]==-1) { rem[t*2]=rem[t]+1; que.push(t*2); } if(t+1<=100000&&rem[t+1]==-1) { rem[t+1]=rem[t]+1; que.push(t+1); } if(t-1>=0&&rem[t-1]==-1) { rem[t-1]=rem[t]+1; que.push(t-1); } } } int main() { ios::sync_with_stdio(false); while(cin>>N>>K) { memset(rem,-1,sizeof(rem)); if(K<=N) cout<<N-K<<endl; else { bfs(); cout<<rem[K]<<endl; } } return 0; }
时间: 2024-10-06 22:54:14