一棵 nn 个节点的树以及 mm 条有向路径.每个点 ii 都有一个权值 wiw_i.如果某条路径包含了 ii 号节点,并且 ii 号节点是该路径上的第 wiw_i 个节点的话就会对ii的答案产生贡献.

求每个点的答案

n,m300000n,m\le 300000

Solution

静下心来好好想一想,其实这是一道并没有那么难的题...

考虑把一条路径拆成两部分,一段向上一段向下

对于点xx来说,一条从sstt,长度为lenilen_i的路径ii要被算贡献需要满足:

  1. xx处于向上的一段中,则depx+wx=depsdep_x + w_x = dep_s
  2. xx处于向下的一段中,则wxdepx=lenideptw_x-dep_x = len_i-dep_t

这两个东西都很好用一个桶维护,通过树上差分求出.具体实现在s/ts/t的位置计算贡献,lcalca的位置消掉贡献即可

表示看网上各种题解都没看懂...

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

inline void proc_status ()
{
ifstream t ("/proc/self/status");
cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

template <typename T> inline int Chkmax (T &a, T b) {return a < b ? a = b, 1 : 0;}
template <typename T> inline int Chkmin (T &a, T b) {return a > b ? a = b, 1 : 0;}

inline int read ()
{
int sum = 0, fl = 1; char ch = getchar();
for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
return sum * fl;
}

const int Maxn = 3e5 + 100;

int N, M;
int e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1], W[Maxn];
pii E[Maxn];

inline void add_edge (int x, int y) { To[++e] = y; Next[e] = Begin[x]; Begin[x] = e; }

int dep[Maxn], anc[22][Maxn];

inline void dfs_pre (int x)
{
dep[x] = dep[anc[0][x]] + 1;
for (int i = 1; i <= 20; ++i) anc[i][x] = anc[i - 1][anc[i - 1][x]];
for (int i = Begin[x]; i; i = Next[i])
{
int y = To[i];
if (y == anc[0][x]) continue;
anc[0][y] = x;
dfs_pre (y);
}
}

inline int get_lca (int x, int y)
{
if (dep[x] < dep[y]) swap(x, y);
for (int i = 20; i >= 0; --i) if (dep[anc[i][x]] >= dep[y]) x = anc[i][x];
if (x == y) return x;
for (int i = 20; i >= 0; --i) if (anc[i][x] != anc[i][y]) x = anc[i][x], y = anc[i][y];
return anc[0][x];
}

vector <int> vec1[Maxn], vec2[Maxn];
int buc[Maxn], Ans[Maxn];

inline void dfs_up (int x)
{
int sum = buc[dep[x] + W[x]];
for (int i = 0; i < vec1[x].size(); ++i) ++buc[vec1[x][i]];
for (int i = Begin[x]; i; i = Next[i])
{
int y = To[i];
if (y == anc[0][x]) continue;
dfs_up (y);
}
Ans[x] += buc[dep[x] + W[x]] - sum;
for (int i = 0; i < vec2[x].size(); ++i) --buc[vec2[x][i]];
}

inline int dis (int x, int y, int k) { return dep[x] + dep[y] - 2 * dep[k]; }

inline void dfs_down (int x)
{
int sum = buc[W[x] - dep[x]];
for (int i = 0; i < vec1[x].size(); ++i) ++buc[vec1[x][i]];
for (int i = Begin[x]; i; i = Next[i])
{
int y = To[i];
if (y == anc[0][x]) continue;
dfs_down (y);
}
for (int i = 0; i < vec2[x].size(); ++i) --buc[vec2[x][i]];
Ans[x] += buc[W[x] - dep[x]] - sum;
}

inline void Solve ()
{
dfs_pre (1);
for (int i = 1; i <= M; ++i)
{
int k = get_lca (E[i].x, E[i].y);
vec1[E[i].x].pb(dep[E[i].x]);
vec2[k].pb(dep[E[i].x]);
}
dfs_up (1);
memset(buc, 0, sizeof buc);
for (int i = 1; i <= N; ++i) vec1[i].clear(), vec2[i].clear();
for (int i = 1; i <= M; ++i)
{
int k = get_lca (E[i].x, E[i].y), sum = dis(E[i].x, E[i].y, k) - dep[E[i].y];
vec1[E[i].y].pb(sum);
vec2[k].pb(sum);
}
dfs_down (1);
for (int i = 1; i <= N; ++i) printf("%d ", Ans[i]);
puts("");
}

inline void Input ()
{
N = read(), M = read();
for (int i = 1; i < N; ++i)
{
int x = read(), y = read();
add_edge (x, y);
add_edge (y, x);
}
for (int i = 1; i <= N; ++i) W[i] = read();
for (int i = 1; i <= M; ++i) E[i].x = read(), E[i].y = read();
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("running.in", "r", stdin);
freopen("running.out", "w", stdout);
#endif
Input();
Solve();
return 0;
}