poj 3107 Godfather 求树的重心【树形dp】

poj 3107 Godfather

和poj 1655差不多，那道会了这个也就差不多了。

题意:从小到大输出树的重心。

题会卡stl，要用邻接表存树。。。。。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int maxn = ;
 const int INF =  << ;
 int head[maxn];
 int son[maxn], ans[maxn];
 bool vis[maxn];
 int n, siz, cnt, tot;
 struct Edge
 {
     int to, next;
 };
 Edge edge[maxn * ];
 
 void Init()
 {
     memset(vis, , sizeof(vis));
     memset(head, -, sizeof(head));
     siz = INF;
     cnt = tot = ;
 }
 
 void add(int u, int v)
 {
     edge[cnt].to = v;
     edge[cnt].next = head[u];
     head[u] = cnt++;
 }
 
 void dfs(int u)
 {
     vis[u] = ;
     son[u] = ;
     int tmp = ;
     for (int i = head[u]; i != -; i = edge[i].next)
     {
         int v = edge[i].to;
         if (!vis[v]) {
             dfs(v);
             son[u] += son[v];
             tmp = max(tmp, son[v]);
         }
     }
     tmp = max(tmp, n - son[u]);
     if (tmp == siz) {
         ans[tot++] = u;
     }
     else if (tmp<siz) {
         tot = ;
         ans[tot++] = u;
         siz = tmp;
     }
 }
 
 int main()
 {
     while (scanf("%d", &n) == )
     {
         Init();
         for (int i = ; i < n; i++) {
             int u, v;
             scanf("%d%d", &u, &v);
             add(u, v);
             add(v, u);
         }
         dfs();
         sort(ans, ans + tot);
         for (int i = ; i<tot; i++) {
             printf("%d ", ans[i]);
         }
         printf("\n");
     }
     return ;
 }