雨中的尾巴

这道题告诉我们，学会开数组大小。

#include
using namespace std;
#define int long long

struct Info
{
    int first, second;
    bool operator<(const Info&that) const
    {
        return first==that.first? second>that.second : first'9') c=='-'&&(n=1), c=getchar();
        while(c>='0'&&c<='9') a=a*10+(c&15), c=getchar();
        n&&(a=-a); return *this;
    }
}in;
const int N = 200005, M = N*16+500; 
const int K = 1e9;
struct Seg
{
    #define lrm int mid = l+r>>1;
    int lc[M], rc[M];
    Info macro[M];
    int stk[M], top, tot;
    int sp()
    {
        if(top)
            return stk[top--];
        else return ++tot;
    }
    void release(int u)
    {
        stk[++top]=u;
        lc[u]=rc[u]=macro[u].first=macro[u].second=0;
    }
    int insert(int u, int l, int r, int pos, int del)
    {
        if(!u)
        {
            if(top)
                u = stk[top--];
            else u = ++tot;
        }
        if(l==r)
        {
            macro[u].first += del;
            macro[u].second = macro[u].first? l:0;
            return u;
        }
        lrm;
        if(pos<=mid)
        {
        //  printf("cuiu%lld, l%lld: ", u, lc[u]);
            lc[u] = insert(lc[u], l, mid, pos, del);
        }
        else
        {
            rc[u] = insert(rc[u], mid+1, r, pos, del);
        }
        macro[u] = max(macro[lc[u]], macro[rc[u]]);
        return u;
    }
    int merge(int u, int v, int l, int r)
    {
        if(!u||!v) return u|v;  //为空，直接返回有的；叶节点的儿子都为空，也为0 
        //有儿子，注意垃圾回收 
        if(l==r)
        {
            macro[u].first += macro[v].first;
            macro[u].second = macro[u].first? l:0;
        }else
        {
            lrm;
            lc[u] = merge(lc[u], lc[v], l, mid);
            rc[u] = merge(rc[u], rc[v], mid+1, r);
            macro[u] = max(macro[lc[u]], macro[rc[u]]);
        }
        release(v); return u;
    }
}seg;
int nxt[N], fst[N], to[N];
void aE(int u, int v)
{
    //d_b(u, v);
    static int i=1; 
    nxt[++i]=fst[u];
    fst[u]=i;
    to[i]=v;
}
int lg, f[N][25], d[N];
void lcainit(int x, int fa)
{
    static queue q;
    q.push(x); d[x]=1;
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        for(int e=fst[u];e;e=nxt[e])
        {
            int v = to[e];
            if(!d[v])
            {
                d[v]=d[u]+1;
                f[v][0]=u;
                q.push(v);
                for(int i=1; i<=lg; i++) f[v][i] = f[f[v][i-1]][i-1];
            }
                
        }
    }
}

int lca(int u, int v)
{
    if(d[u]=d[v]) u=f[u][i];
    if(u==v) return u;
    for(int i=lg; ~i; --i) if(f[u][i]^f[v][i]) u=f[u][i], v=f[v][i];
    return f[u][0];
}
int n,m;
vector qs[N];
int ans[N];
int rt[N];
void dfs(int u)
{
    for(int e=fst[u];e;e=nxt[e])
    {
        int v = to[e];
        if(v==f[u][0]) continue;
        dfs(v);
        rt[u]=seg.merge(rt[u], rt[v], 1, K);
    }
    for(vector::iterator it = qs[u].begin(); it!=qs[u].end(); it++)
    {

        rt[u] = seg.insert(rt[u], 1, K, (*it>0?1:-1)**it, *it>0?1:-1);

    }
    ans[u] = seg.macro[rt[u]].second;
}
signed main()
{
//  freopen("5.in", "r",stdin);
//  freopen("m.txt", "w",stdout);
    in-n-m;
    int x,y,z;
    for(int i=1; i^n; i++)
    {
        in-x-y;
        aE(x,y); aE(y,x);
    }
    lg= log(n)/log(2)+1;
    lcainit(1, 0);
    while(m--)
    {
        in-x-y-z;
        int p = lca(x,y);

        qs[x].push_back(z);
        qs[y].push_back(z);
        qs[p].push_back(-z);
        qs[f[p][0]].push_back(-z);
    }
    dfs(1);
    for(int i=1; i<=n; i++)
        printf("%lld\n", ans[i]);
    return 0;
}