题意：给出一棵树（n个节点)，去掉一个点后，每个子树的节点数都不大于 n/2,求出这样的节点并按升序输出

树形dp，很神圣的样子，就是树形dfs嘛。。。

代码：

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #define nMAX 10010using namespace std;int head[nMAX],dp[nMAX];int s_edge,n,num,ans[nMAX];struct Edge{ int u,v,nxt;}edge[nMAX*2];void addedge(int u,int v){ s_edge++; edge[s_edge].v=v; edge[s_edge].nxt=head[u]; head[u]=s_edge;}int max(int a,int b){ return a>b?a:b;}int dfs(int u,int father){ int cnt=1,SUM=0; for(int e=head[u];e;e=edge[e].nxt) { int v=edge[e].v; if(v==father)continue; int son=dfs(v,u); SUM=max(son,SUM); cnt+=son; } SUM=max(SUM,n-cnt); if(SUM<=n/2){ans[num++]=u;} return cnt;}int main(){ int i,j,k; while(~scanf("%d",&n)) { memset(head,0,sizeof(head)); s_edge=0; for(k=1;k