蓝桥杯备考：BFS之Catch That Cow S

这道题一样，我们可以用BFS层序遍历的方式来做

我们的剪枝策略就是如果走到已经标记过的数，就跳过，因为这条路一定不是最短的

#include <iostream>
#include <cstring>
#include <queue>
const int N = 1e5 + 10;
using namespace std;
int t;
int dist[N];
int x, y;
void bfs()
{
	queue<int>q;
	q.push(x);
	dist[x] = 0;
	while (q.size())
	{
		int t = q.front(); q.pop();
		int a = t + 1;
		int b = t - 1;
		int c = t * 2;
		if (a <= N && dist[a] == -1)
		{
			dist[a] = dist[t] + 1;
			q.push(a);
		}
		if (b >= 0 && dist[b] == -1)
		{
			dist[b] = dist[t] + 1;
			q.push(b);
		}
		if (c <= N && dist[c] == -1)
		{
			dist[c] = dist[t] + 1;
			q.push(c);
		}
		if (a == y || b == y || c == y)
		{
			return;
		}
	}
}
int main()
{
	cin >> t;
	while (t--)
	{
		memset(dist, -1, sizeof(dist));
	    cin >> x >> y;
		bfs();
		cout << dist[y] << endl;
	}





	return 0;
}