CodeForces 101D Castle（树形dp）

题意：给出一棵树，每条边有一个权值，问到达所有节点的路程长度最小平均值是多少。

思路：对于一个结点i，假设它有son[I]棵子树，子树中所有边长两倍是T[I],到达所有节点的距离和的最小值为 dp[I],那么可以得到状态转移方程

dp[I] = sigma(dp[v]) + (son[I]-son[v1])*T[v1]+...+(son[vs])*T[vs-1]

也就是说现在要给子节点规定一个访问顺序，使得这个值最小，观察可以得出可以按照T[I]/son[I]的大小排序。

#include <bits/stdc++.h>
#define eps 1e-6
#define LL long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

const int MAXN = 500000;
//const int MOD = 1e9+7;
//const int INF = 0x3f3f3f3f;
int n;
vector<int> G[MAXN], W[MAXN];
LL dp[MAXN], dsum[MAXN], son[MAXN];
int indice[MAXN];

bool cmp(int u, int v) {
return dsum[u]*son[v] < dsum[v]*son[u];
}

void dfs(int u, int fa) {
dsum[u] = dp[u] = son[u] = 0;
int cnt = 0;
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (v == fa) continue;
dfs(v, u);
dsum[v] += 2*W[u][i];
dsum[u] += dsum[v];
dp[v] += son[v] * W[u][i];
dp[u] += dp[v];
son[u] += son[v];
}
for (int i = 0; i < G[u].size(); i++)
if (G[u][i] == fa)
continue;
else
indice[++cnt] = G[u][i];
sort(indice+1, indice+cnt+1, cmp);
int s = son[u];
for (int i = 1; i <= cnt; i++) {
//if (u == 1) cout << indice[i] << endl;
s -= son[indice[i]];
dp[u] += s*dsum[indice[i]];
}
son[u] += 1;
}

int main()
{
//freopen("input.txt", "r", stdin);
scanf("%d", &n);
for (int i = 1; i < n; i++) {
int u, v, d;
scanf("%d%d%d", &u, &v, &d);
G[u].push_back(v);
G[v].push_back(u);
W[u].push_back(d);
W[v].push_back(d);
}
dfs(1, 0);
//cout << dp[2] << " " << dp[4] << endl;
printf("%.12f", (double)dp[1]/(n-1));
return 0;
}