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 minuteTeleporting: 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?
Line 1: Two space-separated integers: N and K
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
5 17
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.
最小操作問題，典型用bfs，只是bfs 數據量指數增長會超內存，所以我們要想辦法優化。
這個題的有效數據其實就是在[0,2 * K]內，最多2 * K個，所以去除重復數據后的有效數據也就2 * K個，如何去除重復，用bool vis[]記錄是否訪問，后訪問的都是無效的數據，沒必要加入隊列，那么復雜度就是O(K)了
//BFS Memory Limit Exceeded#include<stdio.h>#include<queue>using namespace std;int N,K,ans;int main(){ queue<pair<int,int> > que; while(~scanf("%d%d",&N,&K)){ que.push(make_pair(N,0)); while(true){ int s=que.front().first,t=que.front().second;que.pop(); if(s==K){ 對上面代碼優化后//Accepted 1028kb 32ms#include<stdio.h>#include<string.h>#include<queue>using namespace std;int N,K,ans;queue<pair<int,int> > que;bool vis[200010];int main(){ while(~scanf("%d%d",&N,&K)){ que.push(make_pair(N,0)); memset(vis,0,sizeof(vis)); while(true){ int s=que.front().first,t=que.front().second;que.pop(); if(s==K){ printf("%d/n",t); break; }t++; if(s<K){ if(!vis[s*2]){ que.push(make_pair(s*2,t)); vis[s*2]=true; } if(!vis[s+1]){ que.push(make_pair(s+1,t)); vis[s+1]=true; } } if(s>0&&!vis[s-1]){ que.push(make_pair(s-1,t)); vis[s-1]=true; } } while(!que.empty()) que.pop(); } return 0;}
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