周赛 1009 题解 hdu 2717 Catch That Cow （简单bfs）
Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6016 Accepted Submission(s): 1914
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
4HintThe 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：
你在s点，奶牛在e点，奶牛不动，你每次能从x到x+1或x-1或2*x，要抓住奶牛，问最小步数。
思路：
bfs，利用一个队列，每次出队首nx，按照上述三种方法扩展节点tx，如果tx没有入队列，则入队列，直到e点入队列，这样能保证每一个点都能以最小的步数入队列，就能得到正确答案了。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 100005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int n,m,ans,cnt,tot,flag;
int vis[MAXN]; // 标记数组 并且记录每个点的步数
void bfs()
{
int i,j,t;
queue<int>q; // STL中的队列 当然也可以手写
memset(vis,-1,sizeof(vis)); // 初始化为-1
vis[n]=0; // 起点0步可以到达
q.push(n); // 起点入队
while(!q.empty())
{
int nx=q.front(); // 取出队首
q.pop(); // 队首出队
if(nx==m) // 如果为m 则找到奶牛
{
ans=vis[m];
return ;
}
if(nx>0&&vis[nx-1]==-1) // 扩展tx
{
int tx=nx-1;
vis[tx]=vis[nx]+1;
q.push(tx);
}
if(nx<=100000&&vis[nx+1]==-1) // 扩展tx
{
int tx=nx+1;
vis[tx]=vis[nx]+1;
q.push(tx);
}
if(2*nx<=100000&&vis[2*nx]==-1) // 扩展tx
{
int tx=nx*2;
vis[tx]=vis[nx]+1;
q.push(tx);
}
}
}
int main()
{
int i,j,t;
while(~scanf("%d%d",&n,&m))
{
bfs();
printf("%d\n",ans);
}
return 0;
}