Codeforces 219D. Choosing Capital for Treeland (树dp)

题目链接：http://codeforces.com/contest/219/problem/D

树dp

 1 //#pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <vector>
 8 #include <cmath>
 9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long LL;
15 typedef pair <int, int> P;
16 const int N = 2e5 + 5;
17 struct Edge {
18     int next, to, cost;
19 }edge[N << 1];
20 int head[N], tot, d[N], ans1[N], ans[N];
21 int res[N];
22
23 inline void add_edge(int u, int v, int c) {
24     edge[tot].next = head[u];
25     edge[tot].to = v;
26     edge[tot].cost = c;
27     head[u] = tot++;
28 }
29
30 void dfs1(int u, int p) {
31     d[u] = 1;
32     ans1[u] = 0;
33     for(int i = head[u]; ~i; i = edge[i].next) {
34         int v = edge[i].to;
35         if(v == p)
36             continue;
37         dfs1(v, u);
38         d[u] += d[v];
39         ans1[u] += edge[i].cost + ans1[v];
40     }
41 }
42
43 void dfs2(int u, int p) {
44     ans[u] += ans1[u];
45     for(int i = head[u]; ~i; i = edge[i].next) {
46         int v = edge[i].to;
47         if(v == p)
48             continue;
49         ans[v] = ans[u] - ans1[v] - edge[i].cost + !edge[i].cost;
50         dfs2(v, u);
51     }
52 }
53
54 int main()
55 {
56     memset(head, -1, sizeof(head));
57     int n, u, v;
58     scanf("%d", &n);
59     for(int i = 1; i < n; ++i) {
60         scanf("%d %d", &u, &v);
61         add_edge(u, v, 0);
62         add_edge(v, u, 1);
63     }
64     dfs1(1, -1);
65     dfs2(1, -1);
66     int Min = N, cnt = 0;
67     for(int i = 1; i <= n; ++i) {
68         Min = min(Min, ans[i]);
69     }
70     for(int i = 1; i <= n; ++i) {
71         if(ans[i] == Min) {
72             res[++cnt] = i;
73         }
74     }
75     printf("%d\n", Min);
76     sort(res + 1, res + cnt + 1);
77     for(int i = 1; i <= cnt; ++i) {
78         printf("%d%c", res[i], i == cnt ? ‘\n‘: ‘ ‘);
79     }
80     return 0;
81 }