• 计蒜客模拟赛 #5 (B 题) 动态点分治+线段树


    虽然是裸的换根dp,但是为了在联赛前锻炼码力,强行上了点分树+线段树.   

    写完+调完总共花了不到 $50$ 分钟,感觉还行.  

    code: 

    #include <bits/stdc++.h>   
    #define N 420004  
    #define LL long long  
    #define setIO(s) freopen(s".in","r",stdin) 
    using namespace std;  
    namespace IO 
    {
        char *p1,*p2,buf[100000];
        #define nc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
        int rd() {int x=0; char c=nc(); while(c<48) c=nc(); while(c>47) x=(((x<<2)+x)<<1)+(c^48),c=nc(); return x;}
    };  
    int n,K,edges; 
    int hd[N],nex[N<<1],val[N<<1],to[N<<1],F[N],G[N];      
    inline void add(int u,int v,int c) 
    {
        nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;    
    }   
    struct tree 
    {  
        LL dis[N];  
        int size[N],son[N],top[N],fa[N],dep[N];     
        void dfs1(int u,int ff) 
        { 
            fa[u]=ff,dep[u]=dep[ff]+1,size[u]=1; 
            for(int i=hd[u];i;i=nex[i]) 
            {
                int v=to[i]; 
                if(v==ff) continue;
                dis[v]=dis[u]+val[i];       
                dfs1(v,u);       
                size[u]+=size[v];   
                if(size[v]>size[son[u]])   son[u]=v; 
            }          
        }   
        void dfs2(int u,int tp) 
        {
            top[u]=tp;    
            if(son[u])  dfs2(son[u],tp); 
            for(int i=hd[u];i;i=nex[i]) 
                if(to[i]!=fa[u]&&to[i]!=son[u])    dfs2(to[i],to[i]);  
        }  
        inline int LCA(int x,int y) 
        {
            while(top[x]!=top[y]) 
            {
                dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];  
            } 
            return dep[x]<dep[y]?x:y;  
        } 
        inline LL Dis1(int x,int y) 
        {
            return dis[x]+dis[y]-(dis[LCA(x,y)]<<1);  
        } 
        inline LL Dis2(int x,int y) 
        {
            return dep[x]+dep[y]-(dep[LCA(x,y)]<<1);   
        }
    }tree;  
    struct seg
    {    
        int tot; 
        inline void init() { tot=0;}    
        inline int newnode() { return ++tot; }   
        struct node
        {   
            int ls,rs,s2; 
            LL sum; 
        }t[N*20];     
        void update(int &x,int l,int r,int p,LL v) 
        {
            if(!x) x=newnode();     
            t[x].sum+=v;         
            t[x].s2++;  
            if(l==r) return; 
            int mid=(l+r)>>1;            
            if(p<=mid)    update(t[x].ls,l,mid,p,v);   
            else     update(t[x].rs,mid+1,r,p,v);  
        }   
        LL query(int x,int l,int r,int L,int R) 
        {     
            if(!x) return 0;    
            if(l>=L&&r<=R)    return t[x].sum;  
            int mid=(l+r)>>1;  
            LL re=0ll; 
            if(L<=mid)    re+=query(t[x].ls,l,mid,L,R);                   
            if(R>mid)     re+=query(t[x].rs,mid+1,r,L,R);  
            return re;   
        }
        int q2(int x,int l,int r,int L,int R) 
        {
            if(!x) return 0; 
            if(l>=L&&r<=R)    return t[x].s2;    
            int mid=(l+r)>>1,re=0; 
            if(L<=mid)   re+=q2(t[x].ls,l,mid,L,R);  
            if(R>mid)    re+=q2(t[x].rs,mid+1,r,L,R);  
            return re;   
        }
    }seg; 
    struct opt 
    {   
        int root,sn; 
        int Fa[N],size[N],vis[N],mx[N];    
        void getroot(int u,int ff) 
        {     
            size[u]=1,mx[u]=0;   
            for(int i=hd[u];i;i=nex[i]) 
            {
                int v=to[i]; 
                if(v==ff||vis[v]) continue;  
                getroot(v,u);  
                size[u]+=size[v];            
                mx[u]=max(mx[u], size[v]); 
            }  
            mx[u]=max(mx[u], sn-size[u]);   
            if(mx[u]<mx[root])  root=u;    
        }         
        void calc(int pp,int u,int lst,int ff,int dep,LL w) 
        {          
            if(dep<=K) seg.update(F[pp],0,K,dep,w);    
            if(ff) 
            {
                LL tmp2=tree.Dis1(u,ff);    
                int tmp1=tree.Dis2(u,ff);              
                if(tmp1<=K)     seg.update(G[pp],0,K,tmp1,tmp2);        
            }   
            for(int i=hd[u];i;i=nex[i])   
            {
                int v=to[i];   
                if(vis[v]||v==lst) continue;  
                calc(pp,v,u,ff,dep+1,w+1ll*val[i]);                         
            }
        }  
        void dfs(int u) 
        {    
            vis[u]=1;    
            calc(u,u,0,Fa[u],0,0ll);    
            for(int i=hd[u];i;i=nex[i]) 
            {
                int v=to[i];      
                if(vis[v]) continue;    
                root=0,sn=size[v],getroot(v,u),Fa[root]=u,dfs(root);       
            }
        } 
        LL getdis(int u) 
        {
            LL tmp=seg.query(F[u],0,K,0,K);                             
            int U=u;       
            for(;Fa[u];u=Fa[u])        
            {
                int tt=tree.Dis2(U,Fa[u]);                
                if(tt<=K) 
                {  
                    LL a=seg.query(F[Fa[u]],0,K,0,K-tt)-seg.query(G[u],0,K,0,K-tt);  
                    LL b=seg.q2(F[Fa[u]],0,K,0,K-tt)-seg.q2(G[u],0,K,0,K-tt);  
                    tmp+=a+b*tree.Dis1(U,Fa[u]);          
                }
            }    
            return tmp;   
        }
        void solve() 
        {
            root=0;  
            mx[0]=sn=n;           
            getroot(1,0);       
            dfs(root);        
            int i,j;           
            for(i=1;i<=n;++i) 
            {
                // getdis(i); 
                printf("%lld ",getdis(i));  
            }       
        }
    }opt;  
    int main() 
    {  
        //  setIO("input"); 
        using namespace IO;  
        int i,j;  
        n=rd(),K=rd(); 
        //   scanf("%d%d",&n,&K);   
        for(i=1;i<n;++i) 
        { 
            int x,y,c; 
            x=rd(),y=rd(),c=rd(); 
            add(x,y,c),add(y,x,c); 
        }    
        tree.dfs1(1,0); 
        tree.dfs2(1,1);     
        seg.init();           
        opt.solve();     
        return 0;  
    }
    

      

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  • 原文地址:https://www.cnblogs.com/guangheli/p/11826477.html
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