分析
贪心+并查集。
由于对于一个节点\(i\)，按照贪心的策略，若\(c[i]\)最大，那么我们这一步一定是将其染色，但前提是要将其父节点染色，于是它们便有一层捆绑关系，我们可以将其看做一个节点处理。
代码
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define il inline
#define re register
#define maxn 1005
#define tie0 cin.tie(0),cout.tie(0)
#define fastio ios::sync_with_stdio(false)
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
typedef long long ll;
template <typename T> inline void read(T &x) {
T f = 1; x = 0; char c;
for (c = getchar(); !isdigit(c); c = getchar()) if (c == '-') f = -1;
for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
x *= f;
}
int n, r;
int fa[maxn], f[maxn], c[maxn], val[maxn], times[maxn];
int find(int x) {
return f[x] == x ? x : f[x] = find(f[x]);
}
int main() {
while (scanf("%d%d", &n, &r) && n && r) {
int u, v; c[0] = 0;
memset(val, 0, sizeof(val));
for (int i = 1; i <= n; ++i) read(c[i]);
for (int i = 1; i < n; ++i) {
read(u), read(v);
fa[v] = u;
}
fa[r] = 0;
for (int i = 0; i <= n; ++i) f[i] = i, times[i] = 1;
for (int i = 1; i <= n; ++i) {
u = 0;
for (int j = 1; j <= n; ++j)
if (f[j] == j && (u == 0 || c[u] * times[j] < c[j] * times[u]))
u = j;
v = find(fa[u]);
val[v] += times[v] * c[u] + val[u];
times[v] += times[u], c[v] += c[u];
f[u] = v;
}
printf("%d\n", val[0]);
}
return 0;
}