POJ-3278-Catch That Cow

链接：https://vjudge.net/problem/POJ-3278

题意：

给定n,k。有三种操作n-1,n+1,n*2。

找到从n到k的最少步骤。

思路：

BFS

代码：

#include <iostream>
#include <queue>
using namespace std;
const int MAXN = 1e5+10;
struct Node
{
    int _x;
    int _step;
    Node(int x,int step)
    {
        _x = x;
        _step = step;
    }
};
int vis[MAXN];

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    queue<Node> Q;
    Q.push(Node(n,0));
    vis[n] = 1;
    while (!Q.empty())
    {
        int x = Q.front()._x;
        int step = Q.front()._step;
        if (x == k)
            break;
        if (x-1 >= 0&&x-1<MAXN&&vis[x-1] == 0)
        {
            Q.push(Node(x-1,step+1));
            vis[x-1] = 1;
        }
        if (x+1 >= 0&&x+1<MAXN&&vis[x+1] == 0)
        {
            Q.push(Node(x+1,step+1));
            vis[x+1] = 1;
        }
        if (x*2 >= 0&&x*2<MAXN&&vis[x*2] == 0)
        {
            Q.push(Node(x*2,step+1));
            vis[x*2] = 1;
        }
        Q.pop();
    }
    printf("%d
",Q.front()._step);

    return 0;
}