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

题目传送门

 1 /*
 2     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
 3     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],
 4             此时可以从上个节点的信息递推出来
 5 */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <cmath>
 9 #include <vector>
10 using namespace std;
11
12 const int MAXN = 2e5 + 10;
13 const int INF = 0x3f3f3f3f;
14 struct Edge    {
15     int v, w;
16 };
17 vector<Edge> G[MAXN];
18 int dp[MAXN];
19 bool vis[MAXN];
20 int n, res;
21
22 void DFS(int u)    {
23     vis[u] = true;
24     for (int i=0; i<G[u].size (); ++i)    {
25         int v = G[u][i].v, w = G[u][i].w;
26         if (vis[v])    continue;
27         DFS (v);
28         res += w;
29     }
30 }
31
32 void DFS2(int u)    {
33     vis[u] = true;
34     for (int i=0; i<G[u].size (); ++i)    {
35         int v = G[u][i].v, w = G[u][i].w;
36         if (vis[v])    continue;
37         if (w == 0)    dp[v] = dp[u] + 1;
38         else    dp[v] = dp[u] - 1;
39         DFS2 (v);
40     }
41 }
42
43 void work(void)    {
44     memset (vis, false, sizeof (vis));
45     memset (dp, 0, sizeof (dp));
46
47     res = 0;    DFS (1);    dp[1] = res;
48     memset (vis, false, sizeof (vis));
49     DFS2 (1);
50
51     int mn = INF, p = 0;
52     for (int i=1; i<=n; ++i)    {
53         if (mn >= dp[i])    {
54             mn = dp[i];    p = i;
55         }
56     }
57     printf ("%d\n", mn);
58     for (int i=1; i<=n; ++i)    {
59         if (i == p)    {
60             printf ("%d\n", i);    break;
61         }
62         if (dp[i] == mn)    printf ("%d ", i);
63     }
64 }
65
66 int main(void)    {        //Codeforces Round #135 (Div. 2) D. Choosing Capital for Treeland
67 //    freopen ("A.in", "r", stdin);
68
69     while (scanf ("%d", &n) == 1)    {
70         for (int i=1; i<=n; ++i)    G[i].clear ();
71         for (int i=1; i<=n-1; ++i)    {
72             int u, v;    scanf ("%d%d", &u, &v);
73             G[u].push_back ((Edge) {v, 0});
74             G[v].push_back ((Edge) {u, 1});
75         }
76
77         work ();
78     }
79
80     return 0;
81 }