物流网站开发公司怎么做私人网站

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;
}