[HDU5029][樹鏈剖分][線段樹]Relief grain[好題]

題意：

    給定由n 個村莊構成的一棵樹，進行m 次糧食發放，每次給一條路徑(xi,yi) 上的點發放種類爲zi 的糧食。問m 次發放後，每個村莊領到的數量最多的糧食是哪一種，沒領到糧食輸出0。

題解：

    一看到“樹”和“鏈”，十有八九是樹鏈剖分。
   其實樹鏈剖分也是一種普通的樹形轉線形，只不過它的想法是：我把每條鏈都能剖成一些轉爲線形之後的連續區間。
   所以這道題的做法是，樹鏈剖分後，對於每個重鏈上的區間，在dfs序小的那一個上標記“加上zi ”，在dfs序大的那一個上標記“減去zi ”。最後按dfs序遍歷整棵樹，順便計算答案，途中用線段樹維護每種糧食的個數。

（不要隨便看代碼。不要隨便看代碼。不要隨便看代碼。重要的事情說三遍。）

#ifndef ONLINE_JUDGE
# include "stdafx.h"
# pragma warning(disable:4996)
#endif

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

//Global Variables & Definitions
#define MS(arr, x) memset(arr, x, sizeof(arr))

int N, M;

#define MAXN 100010
#define MAXM 100010
#define MAXE 200020
#define MAXT 400040
#define MAXA 6000030

#define LL 1
#define RR 100010

#define DEFINE_MID int mid = (l + r) >> 1
#define lson (u << 1)
#define rson (u << 1 | 1)
//End Global Variables & Definitions

//Map
struct edge {
    int v, next;
} e[MAXE];

int ecnt;
int h[MAXN];

inline void init_edge() {
    MS(h, ecnt = -1);
}

inline void adde(int u, int v) {
    ++ecnt;
    e[ecnt].v = v;
    e[ecnt].next = h[u];
    h[u] = ecnt;
}
//End Map

//Segment Tree
int maxv[MAXT], maxc[MAXT];

void PushUp(int u) {
    if (!maxv[lson] && !maxv[rson]) {
        maxv[u] = maxc[u] = 0;
    }
    else {
        if (maxv[rson] > maxv[lson]) {
            maxv[u] = maxv[rson];
            maxc[u] = maxc[rson];
        }
        else {
            maxv[u] = maxv[lson];
            maxc[u] = maxc[lson];
        }
    }
}

void Build(int u, int l, int r) {
    maxv[u] = maxc[u] = 0;
    if (l == r) return;

    DEFINE_MID;

    Build(lson, l, mid);
    Build(rson, mid + 1, r);
}

void Change(int u, int l, int r, int p, int v) {
    if (l == r) {
        maxv[u] += v;
        maxc[u] = maxv[u] ? l : 0;
        return;
    }

    DEFINE_MID;

    if (p <= mid) Change(lson, l, mid, p, v);
    else Change(rson, mid + 1, r, p, v);

    PushUp(u);
}

int Query(int u, int l, int r, int p) {
    if (l == r) return maxv[u];

    DEFINE_MID;

    if (p <= mid) return Query(lson, l, mid, p);
    else return Query(rson, mid + 1, r, p);
}
//End Segment Tree

//Tree
int ans[MAXN];
int l[MAXN], r[MAXN], f[MAXN];
int hs[MAXN], ha[MAXN];
int d[MAXN], size[MAXN];
int vis[MAXN];

int p[MAXN], who[MAXN];

int Acth[MAXN][2];

void pre_dfs_A() {
    MS(r, -1);
    MS(vis, 0);
}

void dfs_A(int u, int depth) {
    Acth[u][0] = Acth[u][1] = -1;

    vis[u] = 1;
    d[u] = depth++;

    size[u] = 1;

    int ths = -1, thss = -1, v, ptr = l[u] = -1;
    for (int i = h[u]; ~i; i = e[i].next) if (!vis[v = e[i].v]) {
        if (~ptr) ptr = r[ptr] = v;
        else ptr = l[u] = v;

        f[v] = u;

        dfs_A(v, depth);

        if (size[v] > thss) {
            thss = size[v];
            ths = v;
        }

        size[u] += size[v];
    }

    hs[u] = ths;
}

void pre_dfs_B() {
    ha[1] = 1;
}

void dfs_B(int u, int & s) {
    p[u] = s; who[s] = u; ++s;

    int ths = hs[u];
    if (!~ths) return;
    ha[ths] = ha[u];
    dfs_B(ths, s);

    for (int i = l[u]; ~i; i = r[i]) if (i != ths) {
        ha[i] = i;
        dfs_B(i, s);
    }
}

struct Act {
    int v, next;
} A[MAXA];

int acnt;
inline void init_acts() {
    acnt = -1;
}

inline void adda(int u, int d, int v) {
    ++acnt;
    A[acnt].v = v;
    A[acnt].next = Acth[u][d];
    Acth[u][d] = acnt;
}

void Update(int u, int v, int c) {
    int uf, vf;

    while (u != v) {
        uf = ha[u]; vf = ha[v];

        if (uf == vf) {
            if (d[u] > d[v]) swap(u, v);

            adda(u, 0, c);
            adda(v, 1, c);

            return;
        }
        else {
            if (d[uf] < d[vf]) { swap(uf, vf); swap(u, v); }

            adda(uf, 0, c);
            adda(u, 1, c);

            u = f[uf];
        }
    }

    adda(u, 0, c);
    adda(u, 1, c);
}

void pre_dfs_C() {
    Build(1, LL, RR);
}

void dfs_C(int u) {
    for (int i = Acth[u][0]; ~i; i = A[i].next)
        Change(1, LL, RR, A[i].v, 1);

    ans[u] = maxc[1];

    for (int i = Acth[u][1]; ~i; i = A[i].next)
        Change(1, LL, RR, A[i].v, -1);

    int ths = hs[u];
    if (!~ths) return;

    dfs_C(ths);

    for (int i = l[u]; ~i; i = r[i]) if (i != ths) dfs_C(i);
}
//End Tree

//Main Structure
inline void ir() {
    //Build Tree
    int u, v;
    init_edge();
    for (int i = 1; i < N; ++i) {
        scanf("%d%d", &u, &v);

        adde(u, v); adde(v, u);
    }

    pre_dfs_A();
    dfs_A(1, 0);

    int temp = 1;
    pre_dfs_B();
    dfs_B(1, temp);
}

inline void solve() {
    ir();

    //Deal Acts
    init_acts();

    int u, v, c;
    for (int i = 0; i < M; ++i) {
        scanf("%d%d%d", &u, &v, &c);

        Update(u, v, c);
    }

    pre_dfs_C();
    dfs_C(1);

    for (int i = 1; i <= N; ++i) printf("%d\n", ans[i]);
}

int main() {
    while (scanf("%d%d", &N, &M) == 2 && N) solve();
    return 0;
}