【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+7,inv2=(MOD+1)/2;
      double eps=1e-6;
      int INF=1e9;
      int da[4]={-1,1,0,0};
      int db[4]={0,0,-1,1};


int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,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]+1);
        }
        e=nxt[e];
    }

}

void dfs2(int u,int fa)
{
    int s=0;
    int e=head[u];
    while(e)
    {
        int v=vet[e];
        if(v!=fa)
        {
            s++;
            t1[s]=f[v]+1;
            t2[s]=f[v]+1;
        }
        e=nxt[e];
    }
    rep(i,2,s) t1[i]=t1[i-1]+t1[i];
    per(i,s-1,1) t2[i]=t2[i+1]+t2[i];
    int i=0;
    e=head[u];
    while(e)
    {
        int v=vet[e];
        if(v!=fa)
        {
            i++;
            g[v]=g[u];
            if(i>=2) g[v]=g[v]+t1[i-1];
            if(i<=s-1) g[v]=g[v]+t2[i+1];
            ans[g[v].b+1]=max(ans[g[v].b+1],f[v].b+1);
            ans[f[v].b+1]=max(ans[f[v].b+1],g[v].b+1);
            g[v]=g[v]+1;
        }
        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=0;
        rep(i,1,n) head[i]=0;
        rep(i,1,n)
        {
            f[i].a=f[i].b=0;
            g[i].a=g[i].b=0;
            ans[i]=0;
        }
        rep(i,1,n-1)
        {
            int x=read(),y=read();
            add(x,y);
            add(y,x);
        }
        dfs1(1,0);
        dfs2(1,0);
        per(i,n-1,1) ans[i]=max(ans[i],ans[i+1]);
        ll s=0;
        rep(i,1,n) s+=ans[i];
        printf("%I64d
",s);



    }

    return 0;
}