POJ 2378 Tree Cutting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3949		Accepted: 2375

Description

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

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).

 

求树的重心

CODE：

#include <iostream>
#include <cstdio>
#include <cstring>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)
#define MAX_N 10000 + 10

using namespace std;

int n, u, v;
struct node1{int v, next;} E[MAX_N << 1];
struct node2{int sum, cnt;} N[MAX_N];
int head[MAX_N], top = 0;

void add(int u, int v){
    E[++ top].v = v; E[top].next = head[u]; head[u] = top;
}

int Gao(int x, int last){
    N[x].sum = 1; N[x].cnt = 0;
    for(int i = head[x]; i; i = E[i].next){
        if(E[i].v != last){
            int t = Gao(E[i].v, x);
            N[x].sum += t; N[x].cnt = max(N[x].cnt, t);
        }
    }
    return N[x].sum;
}
 
int main(){
    scanf("%d", &n);
    REP(i, 1, n - 1){
        scanf("%d%d", &u, &v);
        add(u, v); add(v, u);
    }
    Gao(1, 0);
    int res = 99999999;
    REP(i, 1, n){
        N[i].cnt = max(N[i].cnt, n - N[i].sum);
        if(res > N[i].cnt) res = N[i].cnt;
    }
    
    if(res <= n / 2){
        REP(i, 1, n) if(N[i].cnt == res) printf("%d\n", i);
    }
    else printf("NONE\n");
    return 0;
}