标签：stream any sam cout 时间复杂度 奶牛 ati mount ber
题目链接：http://poj.org/problem?id=3278
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?
Input
Output
Sample Input
5 17
Sample Output
4
题意：
有一条数字线段 $[0,100000]$，已知农场主在位置 $N$，逃亡的奶牛在位置 $K$ 不移动，现在农场主在每一分钟都可以有三种行进方式：
往右走一步，往左走一步，传送到为当前位置的两倍的地方；
求最短多少分钟可以抓到奶牛。
题解：
显然是BFS，农场主每行进一次，相当于BFS往下搜一层，那么所有BFS搜索到的位置的深度就相当于行进时间（并且就是到达这个位置的最短时间），
如果某一层遇到了奶牛，就直接终止BFS即可。
时间复杂度：所有的点只会进队一次，$O(N)$。
AC代码：
#include<iostream> #include<cstring> #include<queue> using namespace std; const int maxn=100000; int n,k; int d[maxn+10]; bool vis[maxn+10]; inline bool ok(int x) { if(0<=x && x<=maxn) return 1; else return 0; } void bfs() { queue<int> q; q.push(n); d[n]=0; vis[n]=1; while(!q.empty()) { int now=q.front(); q.pop(); if(now==k) return; int x1=now+1,x2=now-1,x3=2*now; if(ok(x1) && !vis[x1]) q.push(x1),d[x1]=d[now]+1,vis[x1]=1; if(ok(x2) && !vis[x2]) q.push(x2),d[x2]=d[now]+1,vis[x2]=1; if(ok(x3) && !vis[x3]) q.push(x3),d[x3]=d[now]+1,vis[x3]=1; } } int main() { cin>>n>>k; memset(vis,0,sizeof(vis)); bfs(); cout<<d[k]<<endl; }
POJ 3278 - Catch That Cow - [BFS]
标签：stream any sam cout 时间复杂度 奶牛 ati mount ber
原文地址：https://www.cnblogs.com/dilthey/p/9678263.html
