HDU2717 Catch That Cow （BFS）

最新推荐文章于 2021-02-28 08:15:18 发布

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#广度优先算法

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本文介绍了一个有趣的算法问题：一位农夫需要在数轴上通过步行或传送的方式捕捉静止不动的逃逸奶牛。文章详细解释了使用广度优先搜索算法解决此问题的过程，并给出了具体的代码实现。

题目描述

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.

广度优先

用队列储存每一步可能的点，一步一步往下走，直到找到答案

注意标记，少走弯路，保证最短路径

#include <stdlib.h>
#include <stdio.h>
#include <iostream>
#include <queue>
#include <algorithm>
#include <string.h>

using namespace std;

struct node{
    int step;
    int minute;
};
queue<node> que;
int famer,cow;
int track[200000];

void bfs(){
    while(!que.empty()){
        node a=que.front();
        int x=a.step;
        int m=a.minute;
        que.pop();

        if(x==cow){
            cout << m << endl;
            return ;
        }
        if(x>=1&&track[x-1]==0){
            node b;
            b.minute=m+1;
            b.step=x-1;
            track[x-1]=1;
            que.push(b);
        }
        if(x<cow&&track[x*2]==0){  ///有标记，不用担心零重复，减枝
            node b;
            b.minute=m+1;            ///cow10000 x9999 乘以二越界！
            b.step=x*2;
            track[2*x]=1;
            que.push(b);
        }
        if(x<cow&&track[x+1]==0){
            node b;
            b.minute=m+1;
            b.step=x+1;
            track[x+1]=1;
            que.push(b);
        }
    }
}

int main(){
    while(scanf("%d%d",&famer,&cow)!=EOF){
        memset(track,0,sizeof(track));
        while(!que.empty()){
            que.pop();
        }
        node a;
        a.minute=0,a.step=famer;
        track[famer]=1;
        que.push(a);
        bfs();
    }
    return 0;
}