hdu 5029 Relief grain(树链剖分+线段树)

题目链接：hdu 5029 Relief grain

题目大意：给定一棵树，然后每次操作在uv路径上为每一个节点加入一个数w，最后输出每一个节点个数最多的那个数。

解题思路：由于是在树的路径上做操作，所以基本就是树链剖分了。仅仅只是曾经是用一个数组就可以维护值，这题要用

一个vector数组记录。过程中用线段树维护最大值。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 100010;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], s[maxn << 2], d[maxn << 2];

inline void pushup(int u) {
int k = s[lson(u)] < s[rson(u)] ? rson(u) : lson(u);
s[u] = s[k];
d[u] = d[k];
}

void build (int u, int l, int r) {
lc[u] = l;
rc[u] = r;
if (l == r) {
s[u] = 0;
d[u] = l;
return;
}
int mid = (l + r) / 2;
build(lson(u), l, mid);
build(rson(u), mid + 1, r);
pushup(u);
}

void modify(int u, int x, int v) {
if (lc[u] == x && rc[u] == x) {
s[u] += v;
return;
}

int mid = (lc[u] + rc[u]) / 2;
if (x <= mid)
modify(lson(u), x, v);
else
modify(rson(u), x, v);
pushup(u);
}

typedef pair<int,int> pii;
vector<pii> g[maxn];
int N, M, E, first[maxn], jump[maxn * 2], link[maxn * 2], val[maxn];
int id, idx[maxn], dep[maxn], top[maxn], far[maxn], son[maxn], cnt[maxn];

inline void add_Edge (int u, int v) {
link[E] = v;
jump[E] = first[u];
first[u] = E++;
}

void dfs (int u, int pre, int d) {
far[u] = pre;
dep[u] = d;
cnt[u] = 1;
son[u] = 0;

for (int i = first[u]; i + 1; i = jump[i]) {
int v = link[i];
if (v == pre)
continue;
dfs(v, u, d + 1);
cnt[u] += cnt[v];
if (cnt[son[u]] < cnt[v])
son[u] = v;
}
}

void dfs(int u, int rot) {
top[u] = rot;
idx[u] = ++id;
if (son[u])
dfs(son[u], rot);
for (int i = first[u]; i + 1; i = jump[i]) {
int v = link[i];
if (v == far[u] || v == son[u])
continue;
dfs(v, v);
}
}

void init () {
int u, v;
id = E = 0;
memset(first, -1, sizeof(first));
for (int i = 0; i <= N; i++) g[i].clear();

for (int i = 1; i < N; i++) {
scanf("%d%d", &u, &v);
add_Edge(u, v);
add_Edge(v, u);
}
dfs(1, 0, 0);
dfs(1, 1);
}

inline void add(int l, int r, int x) {
g[l].push_back(make_pair(x, 1));
g[r+1].push_back(make_pair(x, -1));
}

void solve (int u, int v, int w) {
int p = top[u], q = top[v];
while (p != q) {
if (dep[p] < dep[q]) {
swap(p, q);
swap(u, v);
}
add(idx[p], idx[u], w);
u = far[p];
p = top[u];
}

if (dep[u] > dep[v])
swap(u, v);
add(idx[u], idx[v], w);
}

int main () {
while (scanf("%d%d", &N, &M) == 2 && N + M) {
init();
int u, v, w;
while (M--) {
scanf("%d%d%d", &u, &v, &w);
solve(u, v, w);
}

int ans = 0;
build(1, 0, 100000);
for (int i = 1; i <= N; i++) {

for (int j = 0; j < g[i].size(); j++)
modify(1, g[i][j].first, g[i][j].second);
val[i] = d[1];
}

for (int i = 1; i <= N; i++)
printf("%d\n", val[idx[i]]);
}
return 0;
}