Doleful Aria

树上倍增lca水题啊，练练手。先一遍dfs搞出树和father[i][j]（意为第i个节点的2^j的祖先）以及distance[i]（意为i到根节点的距离）答案就是distance[x]+distance[y]-2*distance[lca(x,y)]         //lca(x,y)为最近公共祖先代码#include#include#include#define g

方法很多 倍增，线段树，树状数组都是可以的 维护区间最大值和最小值 推荐用倍增 虽然我写了树状数组。。。。#include<cstdio>#include<cstring>#include<algorithm>#define g getchar()#define ll long long#define inf 0x3f3f3f3fusing namespace std;inli