Poj 3278 Catch That Cow

最新推荐文章于 2021-02-27 17:31:42 发布

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本文介绍了一个有趣的算法问题——如何在最短时间内用行走和传送方式捕捉到一只静止不动的逃逸奶牛。通过广度优先搜索策略实现，详细展示了算法的具体实现过程。

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

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;
}