Heoi2014 大工程

题目描述

题解：

一道很水的虚树题。

每次建出虚树后跑一遍树形$dp$，状态比较多。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1000500;
const ll  Inf = 0x3f3f3f3f3f3f3f3fll;
template<typename T>
inline void read(T&x)
{
    T f = 1,c = 0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();}
    x = f*c;
}
int n,q,tl,hed[N],cnt,Hed[N],Cnt,sta[N],Sta[N],sum,Tl,rt;
int use[N],vis[N];
struct EG
{
    int to,nxt,w;
}e[2*N],E[N];
void ae(int f,int t)
{
    e[++cnt].to = t;
    e[cnt].nxt = hed[f];
    hed[f] = cnt;
}
void aE(int f,int t,int w)
{
    E[++Cnt].to = t;
    E[Cnt].nxt = Hed[f];
    E[Cnt].w = w;
    Hed[f] = Cnt;
}
int dep[N],siz[N],top[N],fa[N],son[N],tin[N],tout[N],tim;
void dfs1(int u,int f)
{
    fa[u] = f;
    siz[u] = 1;
    dep[u] = dep[f]+1;
    for(int j=hed[u];j;j=e[j].nxt)
    {
        int to = e[j].to;
        if(to==f)continue;
        dfs1(to,u);
        siz[u]+=siz[to];
        if(siz[to]>siz[son[u]])
            son[u]=to;
    }
}
void dfs2(int u,int Top)
{
    top[u]=Top;
    tin[u]=++tim;
    if(son[u])
    {
        dfs2(son[u],Top);
        for(int j=hed[u];j;j=e[j].nxt)
        {
            int to = e[j].to;
            if(to!=fa[u]&&to!=son[u])
                dfs2(to,to);
        }
    }
    tout[u]=tim;
}
int get_lca(int x,int y)
{
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]])swap(x,y);
        x = fa[top[x]];
    }
    return dep[x]<dep[y]?x:y;
}
bool cmp1(int x,int y){return tin[x]<tin[y];}
bool check(int x,int y){return tin[x]<tin[y]&&tout[x]>=tout[y];}
ll dp[N][6];//siz sum min max
void dfs(int u)
{
    ll minw,maxw;
    if(use[u])
    {
        dp[u][0] = 1;
        dp[u][1] = 0;
        minw = 0;
        dp[u][2] = Inf;
        maxw = 0;
        dp[u][3] = 0;
        dp[u][4] = 0;
        dp[u][5] = 0;
    }else
    {
        dp[u][0] = 0;
        dp[u][1] = 0;
        minw = Inf;
        dp[u][2] = Inf;
        maxw = 0;
        dp[u][3] = 0;
        dp[u][4] = Inf;
        dp[u][5] = 0;
    }
    for(int j=Hed[u];j;j=E[j].nxt)
    {
        int to = E[j].to;
        dfs(to);
        dp[u][0]+=dp[to][0];
        dp[u][1]+=dp[to][1]+E[j].w*dp[to][0]*(sum-dp[to][0]);
        dp[u][2]=min(dp[u][2],min(dp[to][2],minw+dp[to][4]+E[j].w));
        dp[u][3]=max(dp[u][3],max(dp[to][3],maxw+dp[to][5]+E[j].w));
        dp[u][4]=min(dp[u][4],dp[to][4]+E[j].w);
        dp[u][5]=max(dp[u][5],dp[to][5]+E[j].w);
        minw = min(minw,dp[to][4]+E[j].w);
        maxw = max(maxw,dp[to][5]+E[j].w);
    }
}
int main()
{
//  freopen("tt.in","r",stdin);
    read(n);
    for(int u,v,i=1;i<n;i++)
    {
        read(u),read(v);
        ae(u,v),ae(v,u);
    }
    dfs1(1,0),dfs2(1,1);
    read(q);
    for(int i=1;i<=q;i++)
    {
        read(tl);sum=tl;
        for(int x,j=1;j<=tl;j++)
            read(x),sta[j]=x,vis[x]=use[x]=1;
        sort(sta+1,sta+1+tl,cmp1);
        for(int j=1,lim=tl;j<lim;j++)
        {
            int Lca = get_lca(sta[j],sta[j+1]);
            if(!vis[Lca])
                vis[Lca]=1,sta[++tl]=Lca;
        }
        sort(sta+1,sta+1+tl,cmp1);
        for(int j=1;j<=tl;j++)
        {
            while(Tl&&!check(Sta[Tl],sta[j]))Tl--;
            if(!Tl)rt=sta[j];
            else aE(Sta[Tl],sta[j],dep[sta[j]]-dep[Sta[Tl]]);
            Sta[++Tl]=sta[j];
        }
        dfs(rt);
        printf("%lld %lld %lld
",dp[rt][1],dp[rt][2],dp[rt][3]);
        Cnt=Tl=0;
        for(int j=1;j<=tl;j++)
        {
            int v = sta[j];
            use[v]=vis[v]=0;
            Hed[v]=0;
        }
    }
    return 0;
}