Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 76935 | Accepted: 24323 |
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
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.
题意：给出两个数star，end，给出x-1，x+1，x*2四种运算，使star变成end;
思路：直接bfs就好了；
代码：
#include <iostream>
#include <string.h>
#include <queue>
#define MAXN 200000+10
using namespace std; struct Node //***用结构体表示节点，x表示当前值，step记录当前步数
{
int x, step;
}; int star, end, vis[MAXN]; int bfs(int x)
{
Node a, b, c;
a.x = x, a.step = ; //***赋初值
queue<Node>q; //***用队列实现
q.push(a); //***头节点入队
while(!q.empty()) //***如果队列不为空，继续循环
{
b = q.front(); //***取当前队列的头节点做父节点并将其从队列中删除
q.pop();
vis[b.x] = ; //***标记
if(b.x == end) //***如果达到目标，返回步数
{
return b.step;
}
c = b; //***符合条件的儿子入队
if(c.x >= && !vis[c.x-])
{
c.x-=;
c.step++;
vis[c.x]=;
q.push(c);
}
c = b;
if(c.x < end && !vis[c.x+])
{
c.x+=;
c.step++;
vis[c.x]=;
q.push(c);
}
c = b;
if(c.x < end && !vis[*c.x])
{
c.x*=;
c.step++;
vis[c.x]=;
q.push(c);
}
}
return -;
} int main(void)
{
while(cin >> star >> end)
{
if(star >= end) //***如果star>=end，则每次减1，最少需要star-end步
{
cout << star - end << endl;
continue;
}
memset(vis, , sizeof(vis)); //***标记数组清0，不然会对后面的测试产生影响
int ans=bfs(star); //***深搜
cout << ans << endl;
}
return ;
}