P4103 [HEOI2014]大工程

题目

P4103 [HEOI2014]大工程
化简题目：在树上选定(k)个点，求两两路径和，最大的一组路径，最小的一组路径

做法

关键点不多，建个虚树跑一边就好了
(sum_i)为(i)子树各关键点到根节点的距离和，(small_i)为其最小值，(big_i)为其最大值

My complete code

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long LL;
inline LL Read(){
    LL x(0),f(1); char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')f=-1; c=getchar();
    }
    while(c>='0'&&c<='9')
        x=(x<<3)+(x<<1)+c-'0',c=getchar();
    return x*f;
}
const LL maxn=2*1e6+9,inf=0x3f3f3f3f;
LL n,ans1,ans2,ans3;
LL a[maxn],sta[maxn];
struct Tree{
    LL tot,num,head[maxn],dfn[maxn],dep[maxn],size[maxn],key[maxn],inc[maxn][25],fa[maxn];
    LL sum[maxn],big[maxn],small[maxn];
    struct node{
        LL to,next,d;
    }dis[maxn];
    inline void Add(LL u,LL v,LL d){
        dis[++num]=(node){v,head[u],d},head[u]=num;
    }
    void Fdfs(LL u){
        dfn[u]=++tot;
        inc[u][0]=fa[u];
        for(LL i=1;i<=21;++i)
            inc[u][i]=inc[inc[u][i-1]][i-1];
        for(LL i=head[u];i;i=dis[i].next){
            LL v(dis[i].to);
            if(v==fa[u])
                continue;
            fa[v]=u,dep[v]=dep[u]+1,
            Fdfs(v);
        }
    }
    inline LL Lca(LL x,LL y){
        if(dep[x]<dep[y])
            swap(x,y);
        for(LL i=20;i>=0;--i)
            if(dep[inc[x][i]]>=dep[y])
                x=inc[x][i];
        if(x==y)
            return x;
        for(LL i=20;i>=0;--i)
            if(inc[x][i]!=inc[y][i])
                x=inc[x][i],y=inc[y][i];
        return inc[x][0];
    }
    void Dfs(LL u){
        size[u]=key[u],
        sum[u]=big[u]=0,
        small[u]=(key[u]==0)?inf:0;
        for(LL i=head[u];i;i=dis[i].next){
            LL v(dis[i].to),d(dis[i].d);
            Dfs(v);
            if(size[u]){
                ans1+=size[u]*size[v]*d+size[u]*sum[v]+size[v]*sum[u],
                ans2=min(ans2,small[u]+d+small[v]),
                ans3=max(ans3,big[u]+d+big[v]);
            }
            sum[u]+=sum[v]+size[v]*d;
            small[u]=min(small[u],small[v]+d),
            big[u]=max(big[u],big[v]+d),
            size[u]+=size[v];
        }
        head[u]=key[u]=0;
    }
}T,X;
inline bool cmp1(LL x,LL y){
    return T.dfn[x]<T.dfn[y];
}
inline void Solve(){
    LL kase=Read();
    while(kase--){
        LL num(Read());
        for(LL i=1;i<=num;++i)
            a[i]=Read();
        sort(a+1,a+1+num,cmp1);
        LL tp; sta[tp=1]=1;
        for(LL i=1;i<=num;++i){
            X.key[a[i]]=1;
            LL lca=T.Lca(a[i],sta[tp]);
            while(T.dep[sta[tp]]>T.dep[lca])
                if(T.dep[sta[tp-1]]<T.dep[lca])
                    X.Add(lca,sta[tp],T.dep[sta[tp]]-T.dep[lca]),sta[tp]=lca;
                else
                    X.Add(sta[tp-1],sta[tp],T.dep[sta[tp]]-T.dep[sta[tp-1]]),--tp;
            if(sta[tp]!=a[i])
                sta[++tp]=a[i];
        }
        while(tp>1)
            X.Add(sta[tp-1],sta[tp],T.dep[sta[tp]]-T.dep[sta[tp-1]]),--tp;
        ans1=0,ans2=inf,ans3=0;
        X.Dfs(1);
        printf("%lld %lld %lld
",ans1,ans2,ans3);
    }
}
int main(){
    n=Read();
    for(LL i=1;i<n;++i){
        LL u(Read()),v(Read());
        T.Add(u,v,1),T.Add(v,u,1);
    }
    T.Fdfs(1);
    Solve();
    return 0;
}/*
*/