Poj 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 58054		Accepted: 18053

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

思路：既然我们要找位置的，所以必然涉及到当前位置和步数的问题，所以可以设一个结构体，里面有数据步数和位置，然后搜索，当遇到m时，记录下最短时间，当在不断收索时，遇到比当前最优的时间还多的应立即跳出，位置的标记也是优化的一种形式。

#include <iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int n,m;
struct node
{
    int x,ans;
} ;
int vis[1000010],mi;
int jx[]={-1,1};
void bfs()
{
    queue<node>q;
    memset(vis,0,sizeof(vis));
    node f1,f2;
    int i,j,a;
    f1.x=n,f1.ans=0;
    q.push(f1);
    mi=0x3f3f3f3f;
    while(!q.empty())
    {
        f1=q.front();
        q.pop();
        if(f1.x==m)
        {
            if(mi>f1.ans)
            {
                mi=f1.ans;
                continue;
            }
        }
        if(f1.ans>=mi)
            continue;
        for(i=0;i<3;i++)
        {
            f2=f1;
            if(i==2)
                 f2.x=f2.x*2;//这里与i==1||i==0时都要改变f2.x的值，而不是设一个新值，因为还要把此数压入栈内
            else
                f2.x=f2.x+jx[i];
            if(f2.x>=0&&f2.x<=100000&&vis[f2.x]==0)
            {
                vis[f2.x]=1;
                f2.ans++;
                q.push(f2);
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);//注意写上加速语句防TLE
    while(cin>>n>>m)
    {
        if(n==m)
            printf("0\n");
        else
        {
            bfs();
            cout<<mi<<endl;
        }
    }
    return 0;
}