题面:给定一个有$n$个节点的树,每个点又点权$v_i$,每次选取一条树链$[a, b]$,求出$max(v_i - v_j)$,其中$i, j \in [a, b]$且$i$出现在$j$前面,最后树链$[a, b]$上的点点权都加上$v'$
裸的树链剖分,用线段树维护树链。。。查询的时候要用线段树合并。。。然后就没有然后了。。。
代码能力捉鸡QAQQQ,而且貌似是C++程序里面跑的最慢的QAQQQ
/**************************************************************
Problem: 3999
User: rausen
Language: C++
Result: Accepted
Time:4268 ms
Memory:17988 kb
****************************************************************/ #include <cstdio>
#include <algorithm> using namespace std;
typedef long long ll;
const int N = 5e4 + ;
const int inf = 1e9; inline int read(); int n;
int a[N], seq[N], cnt_seq; struct edge {
int next, to;
edge() {}
edge(int _n, int _t) : next(_n), to(_t) {}
} e[N << ]; int first[N], tot; struct tree_node {
int sz, dep, fa, son, v;
int top, pos;
} tr[N]; struct seg {
seg *ls, *rs, *res;
ll mx, mn, mxl, mxr, tag; #define Len (1 << 16)
inline void* operator new(size_t) {
static seg *mempool, *c;
if (mempool == c)
mempool = (c = new seg[Len]) + Len;
return c++;
}
#undef Len
inline seg& operator += (int x) {
mx += x, mn += x, tag += x;
} inline seg* rev() {
swap(mxl, mxr);
return this;
}
inline void update(seg *ls, seg *rs) {
mxl = max(max(ls -> mxl, rs -> mxl), ls -> mx - rs -> mn);
mxr = max(max(ls -> mxr, rs -> mxr), rs -> mx - ls -> mn);
mx = max(ls -> mx, rs -> mx), mn = min(ls -> mn, rs -> mn);
}
inline void push() {
if (tag) {
*ls += tag, *rs += tag;
tag = ;
}
} #define mid (l + r >> 1)
inline void build(int l, int r, int* a) {
res = new()seg;
if (l == r) {
mx = mn = a[l];
return;
}
ls = new()seg(), rs = new()seg;
ls -> build(l, mid, a), rs -> build(mid + , r, a);
update(ls, rs);
} inline void add(int l, int r, int L, int R, int d) {
if (L <= l && r <= R) {
*this += d;
return;
}
push();
if (L <= mid) ls -> add(l, mid, L, R, d);
if (mid < R) rs -> add(mid + , r, L, R, d);
update(ls, rs);
} inline seg* query(int l, int r, int L, int R) {
if (L <= l && r <= R) {
*res = *this;
return res;
}
*res = seg(), push();
if (mid >= R) res = ls -> query(l, mid, L, R);
else if (mid < L) res = rs -> query(mid + , r, L, R);
else res -> update(ls -> query(l, mid, L, R), rs -> query(mid + , r, L, R));
update(ls, rs);
return res;
}
#undef mid
} *T; inline void get(seg *t, int f, int a, int b, int v) {
if (!f) t -> update(T -> query(, n, a, b), t);
else t -> update(t, T -> query(, n, a, b) -> rev());
T -> add(, n, a, b, v);
} inline void pre(seg *t) {
*t = seg();
t -> mx = t -> mxl = t -> mxr = -inf, t -> mn = inf;
} inline void work(int x, int y, int v) {
static seg *left = new()seg, *right = new()seg, *ans = new()seg;
pre(left), pre(right);
while (tr[x].top != tr[y].top) {
if (tr[tr[x].top].dep > tr[tr[y].top].dep)
get(right, , tr[tr[x].top].pos, tr[x].pos, v), x = tr[tr[x].top].fa;
else
get(left, , tr[tr[y].top].pos, tr[y].pos, v), y = tr[tr[y].top].fa;
}
if (tr[x].dep > tr[y].dep) get(right, , tr[y].pos, tr[x].pos, v);
else get(left, , tr[x].pos, tr[y].pos, v);
ans -> update(left, right);
printf("%lld\n", max(ans -> mxl, 0ll));
} inline void Add_Edges(int x, int y) {
e[++tot] = edge(first[x], y), first[x] = tot;
e[++tot] = edge(first[y], x), first[y] = tot;
} #define y e[x].to
void dfs(int p) {
int x;
tr[p].sz = ;
for (x = first[p]; x; x = e[x].next)
if (y != tr[p].fa) {
tr[y].fa = p, tr[y].dep = tr[p].dep + ;
dfs(y);
tr[p].sz += tr[y].sz;
if (!tr[p].son || tr[tr[p].son].sz < tr[y].sz) tr[p].son = y;
}
} void DFS(int p) {
int x;
seq[tr[p].pos = ++cnt_seq] = tr[p].v;
if (!tr[p].son) return;
tr[tr[p].son].top = tr[p].top;
DFS(tr[p].son);
for (x = first[p]; x; x = e[x].next)
if (y != tr[p].fa && y != tr[p].son)
tr[y].top = y, DFS(y);
}
#undef y int main() {
int i, x, y, z, Q;
n = read();
for (i = ; i <= n; ++i) tr[i].v = read();
for (i = ; i < n; ++i)
Add_Edges(read(), read());
dfs();
tr[].top = , DFS();
T = new()seg;
T -> build(, n, seq);
for (Q = read(); Q; --Q) {
x = read(), y = read(), z = read();
work(x, y, z);
}
return ;
} inline int read() {
static int x;
static char ch;
x = , ch = getchar();
while (ch < '' || '' < ch)
ch = getchar();
while ('' <= ch && ch <= '') {
x = x * + ch - '';
ch = getchar();
}
return x;
}