【HDOJ6686】Rikka with Travels（树形DP）

题意：给定一棵n个点，边权为1的树，求有多少个有序数对（l1，l2）使得存在两条互不相交的路径，长度分别为l1和l2

n<=1e5

思路：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 #define N  310000
 #define M  4100000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const ll MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;
       int da[]={-,,,};
       int db[]={,,-,};

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 struct data
 {
     int a,b;
 }f[N],g[N],t1[N],t2[N];

 int tot,ans[N],head[N],vet[N],nxt[N];

 data operator + (const data &a,const data &b)
 {
     return (data){max(a.a,b.a),max(a.a+b.a,max(a.b,b.b))};
 }

 data operator + (const data &a,const int &b)
 {
     return (data){a.a+b,max(a.a+b,a.b)};
 }

 void dfs1(int u,int fa)
 {
     int e=head[u];
     while(e)
     {
         int v=vet[e];
         if(v!=fa)
         {
             dfs1(v,u);
             f[u]=f[u]+(f[v]+);
         }
         e=nxt[e];
     }

 }

 void dfs2(int u,int fa)
 {
     int s=;
     int e=head[u];
     while(e)
     {
         int v=vet[e];
         if(v!=fa)
         {
             s++;
             t1[s]=f[v]+;
             t2[s]=f[v]+;
         }
         e=nxt[e];
     }
     rep(i,,s) t1[i]=t1[i-]+t1[i];
     per(i,s-,) t2[i]=t2[i+]+t2[i];
     int i=;
     e=head[u];
     while(e)
     {
         int v=vet[e];
         if(v!=fa)
         {
             i++;
             g[v]=g[u];
             if(i>=) g[v]=g[v]+t1[i-];
             if(i<=s-) g[v]=g[v]+t2[i+];
             ans[g[v].b+]=max(ans[g[v].b+],f[v].b+);
             ans[f[v].b+]=max(ans[f[v].b+],g[v].b+);
             g[v]=g[v]+;
         }
         e=nxt[e];
     }
     e=head[u];
     while(e)
     {
         int v=vet[e];
         if(v!=fa) dfs2(v,u);
         e=nxt[e];
     }
 }

 void add(int a,int b)
 {
     nxt[++tot]=head[a];
     vet[tot]=b;
     head[a]=tot;
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);

     int cas;
     scanf("%d",&cas);

     while(cas--)
     {
         int n=read();
         tot=;
         rep(i,,n) head[i]=;
         rep(i,,n)
         {
             f[i].a=f[i].b=;
             g[i].a=g[i].b=;
             ans[i]=;
         }
         rep(i,,n-)
         {
             int x=read(),y=read();
             add(x,y);
             add(y,x);
         }
         dfs1(,);
         dfs2(,);
         per(i,n-,) ans[i]=max(ans[i],ans[i+]);
         ll s=;
         rep(i,,n) s+=ans[i];
         printf("%I64d\n",s);

     }

     return ;
 }