「Luogu4556」Vani有约会-雨天的尾巴
传送门
很显然可以考虑树上差分+桶,每次更新一条链就是把这条链上的点在桶对应位置打上 \(1\) 的标记,
最后对每个点取桶中非零值的位置作为答案即可,如果全都是 \(0\) 就输出 \(0\) ,这样的时间复杂度和空间复杂度都是 \(O(nm)\)
考虑优化这一个算法:
我们考虑用权值线段树来代替桶
我们可以考虑用树剖的方式来更新一条链,那么就变成了每次在值域线段树上对 \(\log\) 段区间差分,
需要注意的是,我们把正标记打在深度低的那一边,因为我们可以这样按照 \(\text{dfs}\) 序算答案
小小的证明:
- 对于不贡献当前点的询问,这样做显然是没有影响的
- 对于贡献当前点的询问,它的贡献一定回被重链的顶点打好标记,也一定会被重链底的轻儿子消除
这样一来就得到了一种类似于离线算法的统计答案方法,于是,线段树也就只要开一棵
代码:
#include <cstdio>
#include <vector>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
s = 0; int f = 0; char c = getchar();
while ('0' > c || c > '9') f |= c == '-', c = getchar();
while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
s = f ? -s : s;
}
const int _ = 1e5 + 5;
int tot, head[_], nxt[_ << 1], ver[_ << 1];
inline void Add_edge(int u, int v)
{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v; }
int n, m, ans[_]; vector < int > vec[_];
int dep[_], siz[_], son[_], fa[_];
int dfn[_], rev[_], top[_];
struct node { int mx, pos; } t[_ << 2];
inline int lc(int p) { return p << 1; }
inline int rc(int p) { return p << 1 | 1; }
inline void pushup(int p) {
if (t[lc(p)].mx >= t[rc(p)].mx)
t[p].mx = t[lc(p)].mx, t[p].pos = t[lc(p)].pos;
else
t[p].mx = t[rc(p)].mx, t[p].pos = t[rc(p)].pos;
}
inline void build(int p = 1, int l = 1, int r = 100000) {
if (l == r) { t[p].mx = 0, t[p].pos = l; return ; }
int mid = (l + r) >> 1;
build(lc(p), l, mid), build(rc(p), mid + 1, r), pushup(p);
}
inline void update(int x, int v, int p = 1, int l = 1, int r = 100000) {
if (l == r) { t[p].mx += v; return ; }
int mid = (l + r) >> 1;
if (x <= mid) update(x, v, lc(p), l, mid);
else update(x, v, rc(p), mid + 1, r);
pushup(p);
}
inline void uptRange(int x, int y, int z) {
int fx = top[x], fy = top[y];
while (fx != fy) {
if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
vec[dfn[fx]].push_back(z), vec[dfn[x] + 1].push_back(-z);
x = fa[fx], fx = top[x];
}
if (dep[x] > dep[y]) swap(x, y);
vec[dfn[x]].push_back(z), vec[dfn[y] + 1].push_back(-z);
}
inline void dfs(int u, int f) {
siz[u] = 1, dep[u] = dep[f] + 1, fa[u] = f;
for (rg int i = head[u]; i; i = nxt[i]) {
int v = ver[i]; if (v == f) continue;
dfs(v, u), siz[u] += siz[v];
if (siz[v] > siz[son[u]]) son[u] = v;
}
}
inline void dfs(int u, int f, int topf) {
top[rev[dfn[u] = ++dfn[0]] = u] = topf;
if (son[u]) dfs(son[u], u, topf);
for (rg int i = head[u]; i; i = nxt[i]) {
int v = ver[i]; if (v == f || v == son[u]) continue;
dfs(v, u, v);
}
}
int main() {
#ifndef ONLINE_JUDGE
file("cpp");
#endif
read(n), read(m);
for (rg int u, v, i = 1; i < n; ++i)
read(u), read(v), Add_edge(u, v), Add_edge(v, u);
dfs(1, 0), dfs(1, 0, 1), build();
for (rg int x, y, z; m--; ) read(x), read(y), read(z), uptRange(x, y, z);
for (rg int i = 1; i <= n; ++i) {
for (rg int j : vec[i]) if (j > 0) update(j, 1); else update(-j, -1);
ans[rev[i]] = t[1].mx > 0 ? t[1].pos : 0;
}
for (rg int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
return 0;
}