tree_cuttting(树形dp求解树的重心)

最新推荐文章于 2021-08-10 19:30:55 发布

转载最新推荐文章于 2021-08-10 19:30:55 发布 · 165 阅读

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原文链接：http://www.cnblogs.com/bianjunting/p/11570755.html

在发现贝茜建立昂贵的树状网络后，愤怒的农民约翰决定起诉她。作为报复，贝茜计划切断其中一个谷仓的电源，使网络分裂成更小的部分，每个部分连接的谷仓不超过总数的一半。本篇详细介绍了如何确定哪些谷仓的移除能达成这一目标。

Tree Cutting

After Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible cost, he sued Bessie to mitigate his losses.

Bessie, feeling vindictive, decided to sabotage Farmer John's network by cutting power to one of the barns (thereby disrupting all the connections involving that barn). When Bessie does this, it breaks the network into smaller pieces, each of which retains full connectivity within itself. In order to be as disruptive as possible, Bessie wants to make sure that each of these pieces connects together no more than half the barns on FJ.

Please help Bessie determine all of the barns that would be suitable to disconnect.

Input

* Line 1: A single integer, N. The barns are numbered 1..N.

* Lines 2..N: Each line contains two integers X and Y and represents a connection between barns X and Y.

Output

* Lines 1..?: Each line contains a single integer, the number (from 1..N) of a barn whose removal splits the network into pieces each having at most half the original number of barns. Output the barns in increasing numerical order. If there are no suitable barns, the output should be a single line containing the word "NONE".

Sample Input

Sample Output

3
8

Hint

INPUT DETAILS:

The set of connections in the input describes a "tree": it connects all the barns together and contains no cycles.

OUTPUT DETAILS:

If barn 3 or barn 8 is removed, then the remaining network will have one piece consisting of 5 barns and two pieces containing 2 barns. If any other barn is removed then at least one of the remaining pieces has size at least 6 (which is more than half of the original number of barns, 5).

 1 /*************************************************************************
 2     > File Name: tree_cutting.cpp
 3     > Author: CruelKing
 4     > Mail: 2016586625@qq.com 
 5     > Created Time: 2019年09月22日 星期日 11时31分23秒
 6  ************************************************************************/
 7 
 8 #include <cstdio>
 9 #include <cstring>
10 #include <algorithm>
11 using namespace std;
12 
13 const int maxn = 10000 + 5;
14 
15 int n;
16 
17 int head[maxn], tot;
18 
19 int sonsize[maxn];//统计以i为根节点的子树的大小
20 
21 int dp[maxn];//统计i结点子树中最大的那颗子树的大小
22 
23 int ans;
24 
25     struct Edge {
26     int to, next;
27 } edge[maxn << 1];
28 
29 void init() {
30     memset(head, -1, sizeof head);
31     tot = 0;
32 }
33 
34 void addedge(int u, int v) {
35     edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot ++;
36     edge[tot].to = u;    edge[tot].next = head[v];    head[v] = tot ++;
37 }
38 
39 void dfs(int u, int pre) {
40     sonsize[u] = 1;
41     dp[u] = 0;
42     for(int i = head[u]; ~i; i = edge[i].next) {
43         int v = edge[i].to;
44         if(v == pre) continue;
45         dfs(v, u);
46         sonsize[u] += sonsize[v];
47         dp[u] = max(dp[u], sonsize[v]);
48     }
49     dp[u] = max(dp[u], n - sonsize[u]);
50     ans = min(ans, dp[u]);
51 }
52 
53 int main() {
54     int u, v;
55     init();
56     ans = maxn;
57     scanf("%d", &n);
58     for(int i = 0; i < n - 1; i ++) {
59         scanf("%d %d", &u, &v);
60         addedge(u, v);
61     }
62     dfs(1, 0);
63     int cnt = 0;
64     for(int i = 1; i <= n; i ++) {
65         if(2 * dp[i] <= n) {
66             printf("%d\n", i);
67             cnt ++;
68         }
69     }
70     if(!cnt) printf("NONE\n");
71     return 0;
72 }

转载于:https://www.cnblogs.com/bianjunting/p/11570755.html