标签：des blog http io ar os sp for strong
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 48376 | Accepted: 15149 |
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
Hint
Source
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
int dis[100001],check[100001],n,k,minn=100001;
queue<int> q;
void bfs()
{
q.push(n);
check[n]=1;
while(!q.empty())
{
int x;
x=q.front(),q.pop();
if(x==k)
minn=min(minn,dis[k]);
if(x-1>=0&&x-1<=100000&&!check[x-1])
q.push(x-1),dis[x-1]=dis[x]+1,check[x-1]=1;
if(x+1>=0&&x+1<=100000&&!check[x+1])
q.push(x+1),dis[x+1]=dis[x]+1,check[x+1]=1;
if(x*2>=0&&x*2<=100000&&!check[x*2])
q.push(x*2),dis[x*2]=dis[x]+1,check[x*2]=1;
}
}
int main()
{
memset(dis,0,sizeof(dis));
memset(check,0,sizeof(check));
scanf("%d%d",&n,&k);
bfs();
printf("%d\n",minn);
return 0;
}
标签：des blog http io ar os sp for strong
原文地址：http://www.cnblogs.com/a972290869/p/4101051.html
