BZOJ1509: [NOI2003]逃学的小孩

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复习一下树的直径，多次dfs求解

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#include<bits/stdc++.h>
using namespace std;
int n,m,a,b,ne,c,tpo=1,tp=1,head[405000];
long long ans,d[205000][2],dl[205000];
struct node{int nxt,v;long long dis;}eg[405000];
void adde(int u,int v,int val)
{eg[++ne].v=v;eg[ne].dis=val;eg[ne].nxt=head[u];head[u]=ne;}
void dfs(int u,int fa)
{
    for(int i=head[u];i;i=eg[i].nxt)
    if(eg[i].v!=fa){dl[eg[i].v]=dl[u]+eg[i].dis;dfs(eg[i].v,u);}
}
void dfs2(int u,int fa,int tag)
{
    for(int i=head[u];i;i=eg[i].nxt)
    if(eg[i].v!=fa){d[eg[i].v][tag]=d[u][tag]+eg[i].dis;dfs2(eg[i].v,u,tag);}
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=m;i++)
    {cin>>a>>b>>c;adde(a,b,c);adde(b,a,c);}
    dfs(1,0);
    for(int i=2;i<=n;i++)if(dl[i]>dl[tp])tp=i;
    dl[tp]=0;dfs(tp,0);
    for(int i=2;i<=n;i++)if(dl[i]>dl[tpo])tpo=i;
    dfs2(tp,0,0);dfs2(tpo,0,1);
    for(int i=1;i<=n;i++)ans=max(min(d[i][1],d[i][0]),ans);
    cout<<ans+dl[tpo];
}