題目鏈接
有很多細節上的東西,還是需要寫的細一點,不然很容易wa掉的。
首先,以每個點作爲子樹的根結點來算貢獻很容易發現,所以這裏用一個Dsu的方法,使得的複雜度下降至。
然後dsu的過程一定要注意,要先算答案,再放入子樹的貢獻,當然算答案的時候,我們是要找對應的深度的結點。
於是我們就能算到x的深度爲
但是,我們必須保證才能算它的貢獻,因爲這個等式算出來的deep[x]不保證一定是u的子結點,也有可能會額外的繼承不刪除的dsu部分的貢獻。
是不能保證的,當K比較小的時候deep[id]比較大的時候,容易出事。所以加上判斷。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define eps 1e-8
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N, K, head[maxN], cnt, a[maxN];
struct Eddge
{
int nex, to;
Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN << 1];
inline void addEddge(int u, int v)
{
edge[cnt] = Eddge(head[u], v);
head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
int siz[maxN], dfn[maxN], tot, rid[maxN], Wson[maxN], deep[maxN] = {0};
void pre_dfs(int u, int fa)
{
siz[u] = 1; dfn[u] = ++tot; rid[tot] = u; Wson[u] = -1;
int maxx = 0;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == fa) continue;
deep[v] = deep[u] + 1;
pre_dfs(v, u);
siz[u] += siz[v];
if(maxx < siz[v])
{
maxx = siz[v];
Wson[u] = v;
}
}
}
ll have_deep[maxN << 1] = {0}, num_deep[maxN << 1] = {0}, ans[maxN] = {0};
void dfs(int u, int fa, bool keep)
{
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == fa || v == Wson[u]) continue;
dfs(v, u, false);
}
if(~Wson[u])
{
dfs(Wson[u], u, true);
}
for(int i=head[u], v, id, key; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == fa || v == Wson[u]) continue;
for(int j=dfn[v]; j<dfn[v] + siz[v]; j++)
{
id = rid[j];
key = K + (deep[u] << 1) - deep[id];
if(key > deep[u]) ans[u] += num_deep[key] * a[id] + have_deep[key];
}
for(int j=dfn[v]; j<dfn[v] + siz[v]; j++)
{
id = rid[j];
num_deep[deep[id]]++;
have_deep[deep[id]] += a[id];
}
}
num_deep[deep[u]]++;
have_deep[deep[u]] += a[u];
if(!keep)
{
for(int i=dfn[u], id; i<dfn[u] + siz[u]; i++)
{
id = rid[i];
num_deep[deep[id]]--;
have_deep[deep[id]] -= a[id];
}
}
}
inline void init()
{
cnt = 0;
for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
scanf("%d%d", &N, &K);
init();
for(int i=1; i<=N; i++) scanf("%d", &a[i]);
for(int i=1, u, v; i<N; i++)
{
scanf("%d%d", &u, &v);
_add(u, v);
}
pre_dfs(1, 0);
dfs(1, 0, true);
for(int i=1; i<=N; i++) printf("%lld%c", ans[i], i == N ? '\n' : ' ');
return 0;
}