codeforces 218D Choosing Capital for Treeland 树形DP

给一个有向树每条边可以改变方向求改变的最小边数使其最后连通
先dfs一下求出以1为根的每个点与儿子联通的最小改变数
然后再dfs合并它与父亲的联通信息

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#define maxn 200005
using namespace std;
int last[maxn * 2], Next[maxn * 2], edge[maxn * 2], flag[maxn * 2], l = 0, n;
int am[maxn * 2],  dp[maxn * 2], sym[maxn];
void add(int a, int b, int sym)
{
     edge[l] = b;
     Next[l] = last[a];
     flag[l] = sym;
     last[a] = l;
     l++;
}
void dfs(int pre)
{
     sym[pre] = 1;
     for(int i =last[pre]; i != -1; i = Next[i])
     if(!sym[edge[i]])
     {
         dfs(edge[i]);
         dp[pre] += dp[edge[i]];
         dp[pre] += flag[i];
     }
     sym[pre] = 0;
}
void dfs2(int pre, int cnt)
{
     sym[pre] = 1;
     for(int i =last[pre]; i != -1; i = Next[i])
     if(!sym[edge[i]])
     {
         int CNT = cnt + (dp[pre] - dp[edge[i]]);
         if(flag[i]) CNT--;
         else CNT++;
         dfs2(edge[i], CNT);
     }
     dp[pre] += cnt;
}
int main()
{
    scanf("%d", &n);
    memset(last, -1, sizeof(last));
    for(int i = 1; i < n; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b, 0);
        add(b, a, 1);
    }
    dfs(1);
    dfs2(1, 0);
    int ans = 99999999;
    for(int i = 1; i <= n; i++) ans = min(ans, dp[i]);
    printf("%d\n", ans);
    for(int i = 1; i <= n; i++) if(dp[i] == ans)printf("%d ", i);
    return 0;
}