BFS - 求最短路径

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.

题意：农民有3种走路方式，问怎么走可以最短到达目标位置。

思路： BFS 走感觉还是很好想到的，之前一直有一个地方不是很理解，就是搜素图的时候什么要标记走过的点什么时候不标记走过的点，现在差不多懂点了，每走一步，就把当前这步所有可以到达的位置全部找到，之前走过的点不计入。还有在求步数的搜索题中，特意建一个数组，当前位置的步数等于上一次所走的步数 +1 。

代码：

/*

 * Author:  ry

 * Created Time:  2017/10/26 9:33:23

 * File Name: 2.cpp

 */

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <string>

#include <vector>

#include <stack>

#include <queue>

#include <set>

#include <map>

#include <time.h>

using namespace std;

const int eps = 1e5+5;

const double pi = acos(-1.0);

const int inf = 0x3f3f3f3f;

#define Max(a,b) a>b?a:b

#define Min(a,b) a>b?b:a

#define ll long long

int n, k;

int bu[eps];

void bfs(){

    queue<int>que;

    que.push(n);

    //memset(vis, 0, sizeof(vis));

    memset(bu, -1, sizeof(bu));

    bu[n] = 0;

    while (!que.empty()){

        int a = que.front();

        que.pop();

        if (a == k) break;

        for(int i = 0; i < 3; i++){

            if (i == 0) {

                int b = a - 1;

                if (b >= 0 && b < eps && bu[b] == -1){

                    que.push(b);

                    bu[b] = bu[a] + 1;

                }

            }

            else if (i == 1){

                int b = a + 1;

                if (b >= 0 && b < eps && bu[b] == -1){

                    que.push(b);

                    bu[b] = bu[a] + 1;

                }

            }

            else {

                int b = 2*a;

                if (b >= 0 && b < eps && bu[b] == -1){

                    que.push(b);

                    bu[b] = bu[a] + 1;

                }

            }

        }

    }    

}

int main (){

    while (~scanf("%d%d", &n, &k)){

        bfs();

        //for(int i = 1; i <= k; i++){

            //printf("%d\t", bu[i]);

        //}

        //printf("\n");

        printf("%d\n", bu[k]);

    }

}

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BFS - 求最短路径

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