poj 3278 Catch That Cow(bfs)
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Catch That Cow
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 49503 Accepted: 15506
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK
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
Hint
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.
Source
USACO 2007 Open Silver
题意:
求最小步数使N转换成K。可以:N=N-1,N=N+1,N=N×2;
CODE:
题意:
求最小步数使N转换成K。可以:N=N-1,N=N+1,N=N×2;
CODE:
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<string>#include<algorithm>#include<cstdlib>#include<set>#include<queue>#include<stack>#include<vector>#include<map>#define N 100010#define Mod 10000007#define lson l,mid,idx<<1#define rson mid+1,r,idx<<1|1#define lc idx<<1#define rc idx<<1|1const double EPS = 1e-11;const double PI = acos(-1.0);typedef long long ll;const int INF=1000010;using namespace std;int a[N],b[N];queue<int>mp;int bfs(int k,int n){ mp.push(k); b[k]=1; int mans; while(mp.size()) { mans=mp.front(); mp.pop(); if(mans==n) break; if((mans-1)>=0&&!b[mans-1]) { mp.push(mans-1); a[mans-1]=a[mans]+1; b[mans-1]=1; } if((mans+1)<=100000&&!b[mans+1]) { mp.push(mans+1); a[mans+1]=a[mans]+1; b[mans+1]=1; } if((mans*2)<=100000&&!b[mans*2]) { mp.push(mans*2); a[mans*2]=a[mans]+1; b[mans*2]=1; } } return a[n];}int main(){ int k,n; while(cin>>k>>n) { if(n<=k) printf("%d\n",k-n); else { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); printf("%d\n",bfs(k,n)); } } return 0;}
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