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[ACM] hdu 2717 Catch That Cow (BFS)
Catch That Cow
Problem 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?
* 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
4HintThe 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.
Source
USACO 2007 Open Silver
从位置n到位置k根据规则最少走几步可以到达,规则是从n可以走到n-1,可以走到n+1,也可以走到n*2.用bfs广搜来做,vis[]用来记录是否访问,hash[]用来记录走的步数。
遇到的问题有数组越界,一定要先判断n*2,n-1,n+1是否越界,还有问题就是步数的保存,因为很多节点都是同一个步数下的状态,因为每一步根据规则可以衍生出三种状态。所以用hash[ q.front() 下一个状态]=hash[q.front()]+1.用这种方法来记录步数,最后直接输出hash[k]就可以了。
代码:
#include <iostream> #include <string.h> #include <queue> using namespace std; const int len=100000; int hash[len+10];//hash[i]用来保存走到位置i时是第几步 bool vis[len+10]; int n,k; void bfs() { memset(vis,0,sizeof(vis)); queue<int>q; q.push(n); vis[n]=1; hash[n]=0; while(!q.empty()) { int temp=q.front(); q.pop(); if(temp+1==k||temp-1==k||temp*2==k)//找到 { hash[k]=hash[temp]+1; break; } if(temp-1>=0&&!vis[temp-1]) { q.push(temp-1); vis[temp-1]=1; hash[temp-1]=hash[temp]+1; } if(temp+1<=len&&!vis[temp+1]) { q.push(temp+1); vis[temp+1]=1; hash[temp+1]=hash[temp]+1; } if(temp*2<=len&&!vis[temp*2]) { q.push(temp*2); vis[temp*2]=1; hash[temp*2]=hash[temp]+1; } } } int main() { while(cin>>n>>k) { if(n==k) { cout<<0<<endl; continue; } else bfs(); cout<<hash[k]<<endl; } return 0; }
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