搜索 || BFS || POJ 3278 Catch That Cow

农夫在x位置，下一秒可以到x-1, x+1, 2x，问最少多少步可以到k

*解法：最少步数bfs

要注意的细节蛮多的，写在注释里了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
queue<int> q;
int vis[200005], arr[200005];
void go(int n, int k)
{
    q.push(n);
    vis[n] = 1;
    int flag = 0;
    while(!q.empty())
    {
        if(flag) break;
        int head = q.front(); q.pop();
        int direct[3] = {1, -1, head};
        if(head == k) return;
        for(int i = 0; i < 3; i++)
        {
            int next = head + direct[i];
            if(next >= 0 && next <= 200000 && !vis[next])//坐标一共有1e5但是可以移动到2x 所以next<=2e5；然后next可能小于0，vis[next]直接RE，所以把vis[next]放在最后，先判next>= 0
            {
                q.push(next);
                vis[next] = 1;
                arr[next] = arr[head] + 1;
            }
            if(next == k) flag = 1;
        }
    }
    return;
}
int main()
{
    int n, k;
    while(scanf("%d %d", &n, &k) != EOF)
    {
        while(!q.empty()) q.pop();
        memset(vis, 0, sizeof(vis));
        memset(arr, 0, sizeof(arr));
        go(n, k);
        printf("%d
", arr[k]);
    }
    return 0;
}