Given a tree with n vertices, we want to add an edge between vertex 1 and vertex x, so that the sum of d(1, v) for all vertices v in the tree is minimized, where d(u, v) is the minimum number of edges needed to pass from vertex u to vertex v. Do you know which vertex x we should choose?
Recall that a tree is an undirected connected graph with n vertices and n - 1 edges.
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1 ≤ n ≤ 2 × 105), indicating the number of vertices in the tree.
Each of the following n - 1 lines contains two integers u and v (1 ≤ u, v ≤ n), indicating that there is an edge between vertex u and v in the tree.
It is guaranteed that the given graph is a tree, and the sum of n over all test cases does not exceed 5 × 105. As the stack space of the online judge system is not very large, the maximum depth of the input tree is limited to about 3 × 104.
We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.
Output
For each test case, output a single integer indicating the minimum sum of d(1, v) for all vertices v in the tree (NOT the vertex x you choose).
Sample Input
2 6 1 2 2 3 3 4 3 5 3 6 3 1 2 2 3
Sample Output
8 2
Hint
For the first test case, if we choose x = 3, we will have
d(1, 1) + d(1, 2) + d(1, 3) + d(1, 4) + d(1, 5) + d(1, 6) = 0 + 1 + 1 + 2 + 2 + 2 = 8
It's easy to prove that this is the smallest sum we can achieve.
【分析】给你一棵树，1节点为根。现在在除1号节点外任选一个节点与1节点连一条边，使得其他所有节点到一号节点的距离之和最小，求这个最小的距离之和。
首先咱想一个暴力的方法。直接枚举每一个节点U，使其与1号节点连边，那么U节点及其子树的距离都会被改变，改变值为dis[u]-1（dis[u]=dep[u]-1）.再从1节点到U节点引一条路径，路径上的点及其子树的距离也会被改变，如果该路径上一个节点V的距离改变，那么该节点的所有子树（不包括该路径上的V的儿子节点及其子树）都会被改变，而且改变的差值都是一样的。现在我们就来分析一下哪些点会被更新。记录每一个节点的深度，dep[1]=1.比如dep[u]=6,即1-->2-->3-->4-->5-->6,当加一条边1-->6，则5,6节点及他俩的子树都会被改变，注意这里的路径上的节点都可能有子树，而且受5号节点影响的子树不包括6号节点（前面已说明）。
我们从6节点向上找到最后一个距离会被改变的节点，发现是5节点。而如果在6节点后面再接上7节点呢？可以发现还是5节点，然后再在纸上画几个发现：设向上最后一个被修改的节点为x，则dep[x]=dep[u]/2+2.（U为当前枚举的节点）。再看上边这个例子。5,6节点及其子树将被改变，对于6节点及其子树，改变值为(dis[6]-1)*sz[6],5节点及其子树：(dis[5]-1)*(sz[5]-sz[6]),合并得到 总改变值为2*(sz[5]+sz[6])，推广后，对于当前枚举节点dis是奇数的，差值为（2*（sz[u]+sz[fa[u]]+sz[fa[fa[u]]]...+sz[x]））而且可以推出当前节点dis为偶数时（dep为奇数），改变值为（2*（sz[u]+sz[fa[u]]+sz[fa[fa[u]]]...+sz[x]）-sz[x]）.号公式出来了，现在问题是怎么找这个x节点。嘻嘻，很简单，倍增记录祖先就行了，然后还得记录子树大小前缀和。
#include <bits/stdc++.h> #define inf 1000000000 #define met(a,b) memset(a,b,sizeof a) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define pb push_back #define mp make_pair typedef long long ll; using namespace std; const int N = 2e5+5; const int M = 4e2+50; const ll mod = 1e9+7; int n,m,k; ll ans,sz[N],sum[N],pre; int fa[N][20],dep[N]; vector<int>edg[N]; void dfs1(int u,int f){ sz[u]=1; fa[u][0]=f; dep[u]=dep[f]+1; for(int i=1;i<20;i++){ fa[u][i]=fa[fa[u][i-1]][i-1]; } for(int i=0;i<edg[u].size();i++){ int v=edg[u][i]; if(v==f)continue; dfs1(v,u); sz[u]+=sz[v]; } pre+=dep[u]-1; } void dfs2(int u,int f){ sum[u]=sum[f]+sz[u]; for(int i=0;i<edg[u].size();i++){ int v=edg[u][i]; if(v==f)continue; dfs2(v,u); } } int main() { int op,u,v,x,y; scanf("%d",&op); while(op--){ pre=0; met(fa,0); for(int i=0;i<N;i++){ edg[i].clear(); sum[i]=0; } scanf("%d",&n); for(int i=1;i<n;i++){ scanf("%d%d",&u,&v); edg[u].pb(v); edg[v].pb(u); } dep[0]=0; dfs1(1,0); dfs2(1,0); ans=pre; for(int i=1;i<=n;i++){ int d=dep[i]; int x=d/2+2; if(d==2||d==1)continue; u=i; for(int j=19;j>=0;j--){ if(dep[fa[u][j]]<x)continue; else if(dep[fa[u][j]]>x)u=fa[u][j]; else { u=fa[u][j]; break; } } if(d&1){ v=fa[u][0]; ll ret=2*(sum[i]-sum[v])-sz[u]; ans=min(ans,pre-ret); } else { v=fa[u][0]; ll ret=2*(sum[i]-sum[v]); ans=min(ans,pre-ret); } } printf("%lld ",ans); } return 0; }