[HDU5029][树链剖分][线段树]Relief grain[好题]

题意：

    给定由n个村庄构成的一棵树，进行m次粮食发放，每次给一条路径(xi,yi)上的点发放种类为zi的粮食。问m次发放后，每个村庄领到的数量最多的粮食是哪一种，没领到粮食输出0。

题解：

    一看到“树”和“链”，十有八九是树链剖分。
   其实树链剖分也是一种普通的树形转线形，只不过它的想法是：我把每条链都能剖成一些转为线形之后的连续区间。
   所以这道题的做法是，树链剖分后，对于每个重链上的区间，在dfs序小的那一个上标记“加上zi”，在dfs序大的那一个上标记“减去zi”。最后按dfs序遍历整棵树，顺便计算答案，途中用线段树维护每种粮食的个数。

（不要随便看代码。不要随便看代码。不要随便看代码。重要的事情说三遍。）

#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//Mapstruct 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 Treeint 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//Treeint 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 Structureinline 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;}