poj 2378 树形dp求树的割点

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).

题意
给一颗n个结点的树，节点编号为1~n，把删除一个节点之后，
剩下的分支中节点数量最多的数量不大于总数量一半的编号全部按顺序输出
求切了这个点以后 两个分支的最大数。然后如果切后的结果小于等于 n/2 就可以输出

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;vector<int> v[101010];int dp[101010];int vis[101010];int n;int dfs(int x,int pre){    int sum=1,maxn=0;//求当前结点的子数结点数量（不包括本身，用maxn，总数量用sum，返回的是sum）    for(int i=0;i<v[x].size();i++)    {        if(v[x][i]!=pre)        {            int ans=dfs(v[x][i],x);            maxn=max(maxn,ans);            sum+=ans;        }    }    dp[x]=max(maxn,n-sum);    return sum;}int main(){    while(scanf("%d",&n)!=EOF)    {    for(int i=0;i<n-1;i++)    {        int a,b;        scanf("%d%d",&a,&b);        v[a].push_back(b);        v[b].push_back(a);    }    int cnt=0;    dfs(1,-1);    for(int i=1;i<=n;i++)    {        if(2*dp[i]<=n)        {            printf("%d\n",i );            cnt++;        }    }    if(!cnt) printf("NONE\n");    }}