POJ 3278 Catch That Cow （宽搜）

题目链接：POJ 3278

Describe:

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

题目大意：

FJ和他的牛在一个数轴上，牛不动，FJ每次可以向前一步，或者向后一步，或者跳到当前坐标2倍的位置，问他最少走多少步可以抓到他的牛。

解题思路：

BFS，将每种情况入队，直到找到牛，用dfs会超时。

AC代码：

 

#include <cstdio>
#include <iostream>
#include <queue>
using namespace std;
int x,y,ans,f[200000],p[200000];  // 表示数轴，开两倍大小，p表示走到当前位置时的步数
int main()
{
    scanf("%d%d",&x,&y);
    ans = -1;
    queue<int> q;
    q.push(x); // 起始位置入队
    while(!q.empty())
    {
        int tmp,t = q.front(); q.pop();
        if(t == y) {ans = p[t]; break;}
        for(int i = 0; i < 3; i++) // 枚举三种情况
        {
            if(i == 0) tmp = t-1;
            else if(i == 1) tmp = t+1;
            else if(i == 2) tmp = t*2;
            if(tmp < 0 || tmp > 200000) continue;
            if(f[tmp] == 0) // 若该点没有走过
            {
                f[tmp] = 1; // 标记已经走过
                p[tmp] = p[t]+1;
                q.push(tmp);
            }
        }
    }
    printf("%d",ans);
    return 0;
}

 

小结：

搜索时，要注意不能重复走相同的路。