G - Catch That Cow

最新推荐文章于 2018-07-25 10:38:31 发布

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本文介绍了一种使用广度优先搜索(BFS)算法来解决追击不动目标的问题。在一个数轴上，Farmer John从位置N出发，通过步行或瞬移的方式捕捉位于位置K的静止奶牛。文章详细解释了如何实现这一算法，并给出了具体的示例输入输出。

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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

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.

题目讲让你用广搜
需要注意特殊情况，当famer 和cow在一起时
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#include <iostream>
#include <queue>
#include <string.h>
using namespace std;
bool vis[100002];
int N, K, step;
int dir[2] = {1, -1};
int path[100002];
queue <int> Q;

void bfs(){
    while(!Q.empty()){
        int now = Q.front();
        Q.pop();
        for(int i = 0; i < 3; i++){ //将走的方式看作是一种方向，那么题目就可以转变为BFS
            int tt;
            if(i == 2){
                tt = now * 2;
            }
            else{
                tt = now + dir[i];
            }
            if(tt > 0 && tt < 100001 && !vis[tt]){  //判断方式很简单，只要是在图的范围内并且没有访问过就可以了
                if(tt == K){
                    cout << path[now]+1 << endl;
                    return;
                }
                vis[tt] = true;
                path[tt] = path[now] + 1; //这里用来记录步数,走到第i个点用论文path[i]步，那么从i到下一个点就用了path[i]+1步
                Q.push(tt);
            }
        }
    }
}


int main(){
    ios::sync_with_stdio(false);
    memset(path, 0, sizeof(path));
    memset(vis, false, sizeof(vis));

    cin >> N >> K;
    if(N >= K){ //当cow的位置比john小时候，john只有向后退才会离cow越来越近，只有-1的方式能用，所以直接相减就行了。
        cout << N - K << endl;
    }
    else{
        Q.push(N);
        vis[N] = true;
        bfs();
    }
}