Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 19124 Accepted Submission(s): 5627
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
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
注意点
多组输入
visit数组开大点
#include<bits/stdc++.h>
using namespace std;
struct node{
int now;
int step;
node(){
}
node(int now,int step)
{
this->now = now;
this->step = step;
}
};
int n, k;
int visit[1000005];
int dir[2] = {-1,1};
int bfs()
{
queue<node> q;
q.push(node(n,0));
visit[n] = 1;
while(!q.empty())
{
node t = q.front();
q.pop();
if(t.now == k) return t.step;
for(int i = 0;i < 2;i ++)
{
int tmp = t.now + dir[i];
if(!visit[tmp] && tmp >= 0 && tmp <= 100000)
{
visit[tmp] = 1;
q.push(node(tmp,t.step + 1));
}
}
int tmp = t.now * 2;
if(!visit[tmp] && tmp >= 0 && tmp <= 100000)
{
visit[tmp] = 1;
q.push(node(tmp,t.step + 1));
}
}
return 0;
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(visit,0,sizeof(visit));
cout << bfs() << endl;
}
return 0;
}