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.
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.
* 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
10 1 2 2 3 3 4 4 5 6 7 7 8 8 9 9 10 3 8
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).
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 ", i); 67 cnt ++; 68 } 69 } 70 if(!cnt) printf("NONE "); 71 return 0; 72 }