树形DP Codeforces Round #135 (Div. 2) D. Choosing Capital for Treeland

题目传送门

 /*
     题意：求一个点为根节点，使得到其他所有点的距离最短，是有向边，反向的距离+1
     树形DP：首先假设1为根节点，自下而上计算dp[1](根节点到其他点的距离)，然后再从1开始，自上而下计算dp[v],
             此时可以从上个节点的信息递推出来
 */
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <vector>
 using namespace std;

 const int MAXN = 2e5 + ;
 const int INF = 0x3f3f3f3f;
 struct Edge    {
     int v, w;
 };
 vector<Edge> G[MAXN];
 int dp[MAXN];
 bool vis[MAXN];
 int n, res;

 void DFS(int u)    {
     vis[u] = true;
     for (int i=; i<G[u].size (); ++i)    {
         int v = G[u][i].v, w = G[u][i].w;
         if (vis[v])    continue;
         DFS (v);
         res += w;
     }
 }

 void DFS2(int u)    {
     vis[u] = true;
     for (int i=; i<G[u].size (); ++i)    {
         int v = G[u][i].v, w = G[u][i].w;
         if (vis[v])    continue;
         if (w == )    dp[v] = dp[u] + ;
         else    dp[v] = dp[u] - ;
         DFS2 (v);
     }
 }

 void work(void)    {
     memset (vis, false, sizeof (vis));
     memset (dp, , sizeof (dp));

     res = ;    DFS ();    dp[] = res;
     memset (vis, false, sizeof (vis));
     DFS2 ();

     int mn = INF, p = ;
     for (int i=; i<=n; ++i)    {
         if (mn >= dp[i])    {
             mn = dp[i];    p = i;
         }
     }
     printf ("%d\n", mn);
     for (int i=; i<=n; ++i)    {
         if (i == p)    {
             printf ("%d\n", i);    break;
         }
         if (dp[i] == mn)    printf ("%d ", i);
     }
 }

 int main(void)    {        //Codeforces Round #135 (Div. 2) D. Choosing Capital for Treeland
 //    freopen ("A.in", "r", stdin);

     while (scanf ("%d", &n) == )    {
         for (int i=; i<=n; ++i)    G[i].clear ();
         for (int i=; i<=n-; ++i)    {
             int u, v;    scanf ("%d%d", &u, &v);
             G[u].push_back ((Edge) {v, });
             G[v].push_back ((Edge) {u, });
         }

         work ();
     }

     return ;
 }