Catch That Cow(BFS)

最新推荐文章于 2023-03-23 21:24:14 发布

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最新推荐文章于 2023-03-23 21:24:14 发布

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分类专栏：数据结构-- 图的遍历DFS&BFS

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探讨了在给定的约束条件下，农夫如何通过最优路径在数轴上抓住静止不动的牛，采用广度优先搜索算法解决该问题。

Catch That Cow

Time Limit: 2000 ms Memory Limit: 65536 KiB

Submit Statistic

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?

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

poj3278 有链接提示的题目请先去链接处提交程序，AC后提交到SDUTOJ中，以便查询存档。
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.

Source

汉化：

Catch That Cow

农夫知道一头牛的位置，想要抓住它。农夫和牛都于数轴上，农夫起始位于点 N(0<=N<=100000) ，牛位于点 K(0<=K<=100000) 。农夫有两种移动方式： 1、从 X移动到 X-1或X+1 ，每次移动花费一分钟 2、从 X移动到 2*X ，每次移动花费一分钟假设牛没有意识到农夫的行动，站在原地不。最少要花多少时间才能抓住牛？
Input
一行: 以空格分隔的两个字母: N 和 K
Output
一行: 农夫抓住牛需要的最少时间，单位分钟
Sample Input
5 17
Sample Output
4
Hint
农夫使用最短时间抓住牛的方案如下: 5-10-9-18-17, 需要4分钟.
这道题目就是要自己心中构造一个图，以前做的题目都是二维的图，这次是一个一维的遍历。
然后有三种情况。分别是前进一步，后退一步，前进二倍的步数。

然后是ACcode：

#include <bits/stdc++.h>
using namespace std;
int n, k;
int cat[100005]; // 标记数组
void bfs(int n) // 广度搜索三种情况
{
    cat[n] = 0;
    queue<int> q;
    q.push(n);
    while (!q.empty())
    {
        int num, temp = q.front();
        q.pop();
        /************************************/
        num = temp + 1;
        if (num >= 0 && num <= 100000 && cat[num] == -1)
        {
            cat[num] = cat[temp] + 1;
            q.push(num);
        }
        if (num == k)
            break;
        /************************************/
        num = temp - 1;
        if (num >= 0 && num <= 100000 && cat[num] == -1)
        {
            cat[num] = cat[temp] + 1;
            q.push(num);
        }
        if (num == k)
            break;
        /************************************/
        num = 2 * temp;
        if (num >= 0 && num <= 100000 && cat[num] == -1)
        {
            cat[num] = cat[temp] + 1;
            q.push(num);
        }
        if (num == k)
            break;
    }
    cout << cat[k] << endl;
}
int main()
{
    ios::sync_with_stdio(false);
    while (cin >> n >> k)
    {
        memset(cat, -1, sizeof(cat));

        if (n > k)
            cout << n - k << endl; // 剪枝
        else
            bfs(n);
    }
    return 0;
}


/***************************************************
User name: QXQZX
Result: Accepted
Take time: 0ms
Take Memory: 604KB
Submit time: 2019-01-06 23:03:12
****************************************************/