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(简单) POJ 1278 Catch That Cow，回溯。
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;}
(简单) POJ 1278 Catch That Cow，回溯。
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