B - Edge to the Root （树上dfs+思维）

Posted 2021-02-16 zhangbuang

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++.

<h4< dd="">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).

<h4< dd="">Sample Input

2
6
1 2
2 3
3 4
3 5
3 6
3
1 2
2 3

<h4< dd="">Sample Output

8
2

<h4< dd="">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.

 

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这题有人把它分在了树形dp里，感觉并不像是dp，有点从上到下递推的意思，

开始知道是从上往下推，但是就是想不出来是怎么推了，这就很蒟了，

其实就是考虑我把这个边往下移能带来什么后果，

比如从f 转移到了f的儿子son

son整个子树所经过的距离全都少了1

然后f和root中点 到 f 间所有的点的距离都增加了1

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 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define rep(i, a, b)    for (int i(a); i <= (b); ++i)
 6 #define dec(i, a, b)    for (int i(a); i >= (b); --i)
 7 
 8 typedef long long LL;
 9 
10 const int N = 2e5 + 10;
11 
12 int T, n;
13 int sz[N], deep[N];
14 int c[N];
15 
16 LL  f[N];
17 LL  ans[N], all, ret;
18 vector <int> v[N];
19 
20 void dfs(int x, int fa, int dep){
21     sz[x]   = 1;
22     f[x]    = 0;
23     deep[x] = dep;
24 
25     for (int i=0;i<v[x].size();i++)  {
26         int u=v[x][i];
27         if (u == fa) continue;
28         dfs(u, x, dep + 1);
29         sz[x] += sz[u];
30         f[x]  += 0ll + f[u] + sz[u];
31     }
32 }
33 
34 void solve(int x, int fa, int dep)
35 {
36       for (int i=0;i<v[x].size();i++) 
37      {
38         int u=v[x][i];
39         if (u == fa) continue;
40         c[dep] = u;
41         if (deep[u] >= 2) ans[u] = ans[x] + sz[c[dep / 2 + 1]] - 2 * sz[u];
42         else ans[u] = ans[x];
43         solve(u, x, dep + 1);
44     }
45 }
46         
47 
48 int main(){
49 
50     scanf("%d", &T);
51 
52     while (T--){
53         scanf("%d", &n);
54         rep(i, 0, n + 1) v[i].clear();
55         rep(i, 2, n){
56             int x, y;
57             scanf("%d%d", &x, &y);
58             v[x].push_back(y);
59             v[y].push_back(x);
60         }
61 
62         dfs(1, 0, 0);
63         ans[1] = f[1];
64         c[0] = 1;
65 
66         solve(1, 0, 1);
67 
68         ret = ans[1];
69 
70         rep(i, 2, n) ret = min(ret, ans[i]);
71         printf("%lld
", ret);
72     }
73 
74 
75     return 0;
76 }