#include
      
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            using namespace std;const int maxn = 2*1e5+100;int dp[maxn]; //dp[i]表示以i为根遍历全部的点的最小的权值。struct Edge{ int u,v,w; int next; Edge(int u=0,int v=0,int w=0,int next=0): u(u), v(v), w(w), next(next) {}}edges[maxn*2];int head[maxn],cnt;int sum; //以1为根的树至少要修改的边数。int ans;void addedge(int u,int v){ edges[cnt] = Edge(u,v,0,head[u]); head[u] = cnt++; edges[cnt] = Edge(v,u,1,head[v]); head[v] = cnt++;}void dfs1(int u,int fa){ for(int i=head[u];i!=-1;i=edges[i].next){ Edge& e = edges[i]; if(e.v == fa) continue; sum += e.w; dfs1(e.v,u); }}void dfs2(int u,int fa){ for(int i=head[u];i!=-1;i=edges[i].next){ Edge& e = edges[i]; if(e.v == fa) continue; if(e.w == 1) dp[e.v] = dp[u] - 1; //由于父亲节点的dp[u]已经算出，只需要更新就是了。 else dp[e.v] = dp[u] + 1; ans = min(ans,dp[e.v]); dfs2(e.v,u); }}int main(){ //freopen("E:\\acm\\input.txt","r",stdin); int n; cin>>n; memset(head,-1,sizeof(head)); cnt = 0; for(int i=1;i