poj3278——bfs
POJ 3278 对数轴进行一维bfs
Catch That Cow
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 52161 | Accepted: 16355 |
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 pointK (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 17
Sample Output
4
题意:在数轴上求规定走法的最短路
思路:bfs
很基础的一个bfs,一开始由于边界没控制好RE了好多次,最后发现是处理边界的时候由于语句顺序不对导致vis数组越界了,最终还是AC了过去
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
//poj3278_bfs #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std; const int maxn=100020; int n,k; bool vis[maxn]; struct node { int pos,step; }; int bfs() { memset(vis,0,sizeof(vis)); queue<node> q; q.push({n,0}); vis[n]=1; while(!q.empty()){ node now=q.front(); q.pop(); if(now.pos==k) return now.step; int next; //move to now.pos-1 next=now.pos-1; if(next>=0&&next<maxn&&!vis[next]){ //控制边界时注意要将vis放在next<maxn之后,避免数组越界 q.push({next,now.step+1}); vis[next]=1; } //move to now.pos+1 next=now.pos+1; if(next<maxn&&!vis[next]){ q.push({next,now.step+1}); vis[next]=1; } //move to now.pos*2 next=now.pos*2; if(next<maxn&&!vis[next]&&next!=0){ q.push({next,now.step+1}); vis[next]=1; } } return false; } int main() { while(cin>>n>>k){ cout<<bfs()<<endl; } return 0; }
没有AC不了的题,只有不努力的ACMER!