POJ2378(树形DP)

Tree Cutting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4156		Accepted: 2512

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

题意：问删除哪些结点后，剩余各个子树的大小均不超过总结点数目的一半。
树形DP的核心：将有向树当作无向树，从根节点结点遍历到叶子节点，从叶子节点回缩到根节点。
思路：一次DFS可求出每个结点的最大子树(不包含其本身)maxn以及子树之和sum(包含其本身),若(n-sum)<=n/2&&maxn<=n/2,则为答案。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN=10005;
vector<int> tree[MAXN];
int n;
int mx[MAXN];
int sum[MAXN];
void dfs(int u,int fa)
{
    sum[u]=1;
    int maxn=0;
    int s=0;
    for(int i=0;i<tree[u].size();i++)
    {
        int v=tree[u][i];
        if(v==fa)
            continue;
        dfs(v,u);
        s+=sum[v];
        maxn=max(maxn,sum[v]);
    }
    mx[u]=maxn;
    sum[u]+=s;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(sum,0,sizeof(sum));
        memset(mx,0,sizeof(mx));
        for(int i=1;i<=n;i++)
            tree[i].clear();
        
        for(int i=1;i<=n-1;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            tree[u].push_back(v);
            tree[v].push_back(u);
        }
        dfs(1,-1);
        bool flag=false;
        for(int i=1;i<=n;i++)
        {
            if(max(n-sum[i],mx[i])<=n/2)
            {
                printf("%d
",i);
                flag=true;
            }
        }
        if(!flag)    printf("NONE
");
    }
    return 0;
}