http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5568
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.
Author: WENG, Caizhi
Source: The 17th Zhejiang University Programming Contest Sponsored by TuSimple
题目大意：给你一棵树，这棵树的每条边的length都是1。然后这棵树是以1为root，定义他的weight就是所有的节点的deep和。
现在你有一个操作，对于任意的u，可以从(1,u)连接一条边，问选择哪个节点u连接可以使得树的weight最小。
思路：其实很早就有思路了...但是最近心情太烦了，写的时候都很烦，就没有做。
刚开始sb的用线段树啊，树状数组啊来维护，发现好烦。然后第二天删光了所有的代码重新来，但还是摆脱不了树状数组...（感觉就算写对了应该也是TLE了）
今天突然发现可以维护一下前缀就好了（真的是......）
首先我们可以发现，对于(1,u)连接了边以后，那么深度为deep[u]/2 + 1的这些点的deep都会发生改变。
我们从1开始进行dfs，然后我们对于经过的所有的节点，都把他定义为deep = 1.然后定义前面一个子树的区间为[l, r]，当前的区间为[L, R]，然后利用这个进行修改即可
具体的就是容斥一下就好了，还不懂看着代码里面的注释，然后自己画一画
//看看会不会爆int!数组会不会少了一维！ //取物问题一定要小心先手胜利的条件 #include <bits/stdc++.h> using namespace std; #pragma comment(linker,"/STACK:102400000,102400000") #define LL long long #define ALL(a) a.begin(), a.end() #define pb push_back #define mk make_pair #define fi first #define se second #define haha printf("haha ") const int maxn = 2e5 + 5; int n; vector<int> G[maxn]; int deep[maxn], sz[maxn]; LL cnt[maxn]; void dfs_sz(int u, int fa, int d){ deep[u] = d; sz[u] = 1; cnt[u] = d; for (int i = 0; i < G[u].size(); i++){ int v = G[u][i]; if (v == fa) continue; dfs_sz(v, u, d + 1); sz[u] += sz[v]; cnt[u] += cnt[v]; } } LL ans; LL rest, res, addval; ///rest表示每次需要加回来的东西是多少，addval表示rest的和 ///res表示把路上经过的点deep都变成1所剩下的val LL pre[maxn];//表示深度为l的有多少节点是被修改了的 void dfs_solve(int u, int fa, int l, int r){ pre[deep[u]] = sz[u]; int L = deep[u] / 2 + 1, R = deep[u]; if (R > r) res -= sz[u] * (deep[u] - 1);///减去子树的 if (L > l) rest -= pre[l];///减去之前子树的size addval += rest; ans = min(ans, addval + res); for (int i = 0; i < G[u].size(); i++){ int v = G[u][i]; if (v == fa) continue; pre[deep[u]] -= sz[v]; rest += pre[deep[u]]; res += (deep[u] - 1) * sz[v]; dfs_solve(v, u, L, R); res -= (deep[u] - 1) * sz[v]; rest -= pre[deep[u]]; pre[deep[u]] += sz[v]; } if (R > r) res += sz[u] * (deep[u] - 1); addval -= rest; if (L > l) rest += pre[l]; } int main(){ int t; cin >> t; while (t--){ scanf("%d", &n); for (int i = 1; i <= n; i++) G[i].clear(); for (int i = 1; i < n; i++){ int u, v; scanf("%d%d", &u, &v); G[u].pb(v), G[v].pb(u); } dfs_sz(1, 0, 0); res = ans = cnt[1]; addval = 0; for (int i = 0; i < G[1].size(); i++){ int v = G[1][i]; dfs_solve(v, 1, 1, 0); } printf("%lld ", ans); } return 0; } /* 456 7 1 2 2 3 3 4 3 5 5 7 4 6 ans = 12 */