7.12 POJ 3278 A - 广搜基础

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本文介绍了一道经典的广度优先搜索(BFS)算法题目——如何最短时间找到静止不动的逃跑奶牛。FJ从起点出发，可以选择步行到相邻位置或瞬移到两倍当前位置。文章提供了一个C++实现示例，展示了如何使用队列来解决此类问题。

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A - 广搜基础

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3278

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.

每个节点处有三种走法：+1， -1和*2；

#include<iostream>
#include<queue>
using namespace std;
const int MAXN = 100001;
bool visit[MAXN];
int step[MAXN];
queue<int> que;
int bfs(int n, int k)
{
    int head, next;
    que.push(n);
    visit[n] = true;
    step[n] == 0;
    while(!que.empty())
    {
        head = que.front();
        que.pop();
        for(int i = 0; i < 3; i++)
        {
            if(i == 0) {next = head + 1;}
            else if(i == 1){next = head - 1;}
            else next = head * 2;
            if (next > MAXN || next < 0) continue;
            if(!visit[next])
            {
                que.push(next);
                step[next] = step[head] + 1;
                visit[next] = true;
            }
            if(next == k) {return step[next]; }
        }
    }
}
int main()
{
    int n, k;
    cin >> n >> k;
    cout << bfs(n, k) << endl;
    return 0;
}