| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 40350 | Accepted: 12560 |
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
题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1；2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次可以移动到牛所在的位置。所以可以用广搜来搜索这三个状态，直到搜索到牛所在的位置。
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int N = 200100;
int n, k;
struct node
{
int x, step;
};
queue<node> Q;
int vis[N];
void BFS()
{ //搜索关于X的三种状态
int X, STEP;
while(!Q.empty())
{
node tmp = Q.front();
Q.pop();
X = tmp.x;
STEP = tmp.step;
if(X == k)
{
printf("%d\n",STEP);
return ;
}
if(X >= 1 && !vis[X - 1]) //要保证减1后有意义,所以要X >= 1
{
node temp;
vis[X - 1] = 1;
temp.x = X - 1;
temp.step = STEP + 1;
Q.push(temp);
}
if(X <= k && !vis[X + 1])
{
node temp;
vis[X + 1] = 1;
temp.x = X + 1;
temp.step = STEP + 1;
Q.push(temp);
}
if(X <= k && !vis[X * 2])
{
node temp;
vis[X * 2] = 1;
temp.x = 2 * X;
temp.step = STEP + 1;
Q.push(temp);
}
}
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
while(!Q.empty()) Q.pop();
memset(vis,0,sizeof(vis));
vis[n] = 1;
node t;
t.x = n, t.step = 0;
Q.push(t);
BFS();
}
return 0;
}