POJ 3278 Catch That Cow

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最新推荐文章于 2024-09-04 11:33:10 发布

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分类专栏：搜索

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本文介绍了一个简单的搜索问题——如何在最短时间内找到并捕捉到一只静止不动的逃逸奶牛。通过使用广度优先搜索算法，文章提供了一种有效的解决方案，并详细展示了其代码实现过程。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions:106274		Accepted: 33215

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

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.

题解：暴力找一找。
代码：
#include<cstdio>
#include<queue>
#include<string.h>
using namespace std;
const int maxn=200001;
bool vis[maxn];
struct node
{
    int id,time;
};
queue<node>P;
int main()
{
    int n,k;scanf("%d%d",&n,&k);
    memset(vis,0,sizeof(vis));
    node r;r.id=n;r.time=0;
    vis[n]=1;
    P.push(r);
    while(!P.empty())
    {
        r=P.front();P.pop();
        if(r.id==k)
        {
            printf("%d\n",r.time);
            break;
        }
        int x=r.id+1,y=r.id-1,z=r.id*2;
        node e;
        if(x>=0&&x<maxn&&!vis[x])
        {
            e.id=x;e.time=r.time+1;
            vis[x]=1;
            P.push(e);
        }
        if(y>=0&&y<maxn&&!vis[y])
        {
            e.id=y;e.time=r.time+1;
            vis[y]=1;
            P.push(e);
        }
        if(z<maxn&&!vis[z])
        {
            e.id=z;e.time=r.time+1;
            vis[z]=1;
            P.push(e);
        }
    }
    return 0;
}