[HDU5029]Relief grain(树链剖分+线段树)

两种显然的做法：

1.树上差分＋线段树合并。$O(nlog n)$

2.树链剖分转为序列上差分＋线段树。$O(nlog^2 n)$

后一种的代码：

#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ls (x<<1)
#define rs (ls|1)
#define lson ls,L,mid
#define rson rs,mid+1,R
#define mem(a) memset(a,0,sizeof(a))
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
#define For(i,x) for (int i=h[x],k; i; i=nxt[i])
typedef long long ll;
using namespace std;

const int N=100010;
int n,m,u,v,w,tim,cnt,id[N<<2],s[N<<2],to[N<<1],nxt[N<<1];
int son[N],top[N],pos[N],d[N],sz[N],ans[N],fa[N],h[N];
vector<int>V[N];
void add(int u,int v){ to[++cnt]=v; nxt[cnt]=h[u]; h[u]=cnt; }
void init(){
    cnt=0; tim=0;
    rep(i,1,n) h[i]=son[i]=0,V[i].clear();
}

void dfs1(int x){
    sz[x]=1; d[x]=d[fa[x]]+1;
    For(i,x) if ((k=to[i])!=fa[x]){
        fa[k]=x; dfs1(k); sz[x]+=sz[k];
        if (sz[k]>sz[son[x]]) son[x]=k;
    }
}

void dfs2(int x,int z){
    pos[x]=++tim; top[x]=z;
    if (son[x]) dfs2(son[x],z);
    For(i,x) if ((k=to[i])!=fa[x] && k!=son[x]) dfs2(k,k);
}

void work(int l,int r,int c){ V[l].push_back(c),V[r+1].push_back(-c); }

void solve(int u,int v,int c){
    for (; top[u]!=top[v]; u=fa[top[u]]){
        if (d[top[u]]<d[top[v]]) swap(u,v);
        work(pos[top[u]],pos[u],c);
    }
    if (d[u]>d[v]) swap(u,v);
    work(pos[u],pos[v],c);
}

void upd(int x){
    int k=s[ls]<s[rs] ? rs : ls;
    s[x]=s[k]; id[x]=id[k];
}

void build(int x,int L,int R){
    if (L==R){ s[x]=0; id[x]=L; return; }
    int mid=(L+R)>>1;
    build(lson); build(rson); upd(x);
}

void ins(int x,int L,int R,int pos,int k){
    if (L==R){ s[x]+=k; return; }
    int mid=(L+R)>>1;
    if (pos<=mid) ins(lson,pos,k); else ins(rson,pos,k);
    upd(x);
}

int main(){
    freopen("hdu5029.in","r",stdin);
    freopen("hdu5029.out","w",stdout);
    while (scanf("%d%d",&n,&m)==2 && n+m){
        init(); int mx=0;
        rep(i,2,n) scanf("%d%d",&u,&v),add(u,v),add(v,u);
        fa[1]=0; dfs1(1); dfs2(1,1);
        rep(i,1,m) scanf("%d%d%d",&u,&v,&w),solve(u,v,w),mx=max(mx,w);
        build(1,0,mx);
        rep(i,1,n){
            rep(j,0,(int)V[i].size()-1)
                if (V[i][j]>0) ins(1,0,mx,V[i][j],1); else ins(1,0,mx,-V[i][j],-1);
            ans[i]=id[1];
        }
        rep(i,1,n) printf("%d
",ans[pos[i]]);
    }
    return 0;
}