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。
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
#include
#include
#include
#include
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;
}