树形DP Gym 100496H House of Representatives

 

题目传送门

/*
    题意：寻找一个根节点，求min f(u) = ∑ρ(v, u) * p(v)。ρ(v, u)是u到v的距离，p(v)是v点的权值
    树形DP：先从1出发遍历第一次，sum[u]计算u到所有子节点v的路径权值(之后的点路径有叠加，所以先把路径权值加后*w)，
            计算f[u](缺少u节点以上的信息)。然后再遍历一遍，之前是DFS从下往上逆推，现在是顺推，把u节点以上的信息加上
    dp的部分不是很多，两个DFS函数想了很久，还是没完全理解:(
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

typedef long long ll;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int p[MAXN];
ll f[MAXN];
ll sum[MAXN];
vector<pair<int, int> > G[MAXN];

void DFS1(int u, int rt)
{
    f[u] = 0;    sum[u] = p[u];
    for (int i=0; i<G[u].size (); ++i)
    {
        int v = G[u][i].first;    int w = G[u][i].second;
        if (v == rt)    continue;
        DFS1 (v, u);
        f[u] += f[v] + sum[v] * w;    sum[u] += sum[v];
    }
}

void DFS2(int u, int rt)
{
    for (int i=0; i<G[u].size (); ++i)
    {
        int v = G[u][i].first;    int w = G[u][i].second;
        if (v == rtGym 100496H House of Representatives)    continue;
        f[v] = f[u] - sum[v] * w + (sum[u] - sum[v]) * w;
        sum[v] = sum[u];
        DFS2 (v, u);
    }
}

int main(void)        //Gym 100496H House of Representatives
{
//    freopen ("H.in", "r", stdin);
    freopen ("house.in", "r", stdin);
    freopen ("house.out", "w", stdout);

    int n;
    while (scanf ("%d", &n) == 1)
    {
        for (int i=1; i<=n; ++i)    scanf ("%d", &p[i]);
        for (int i=1; i<=n; ++i)    G[i].clear ();

        for (int i=1; i<=n-1; ++i)
        {
            int u, v, w;    scanf ("%d%d%d", &u, &v, &w);
            G[u].push_back (make_pair (v, w));
            G[v].push_back (make_pair (u, w));
        }

        DFS1 (1, -1);    DFS2 (1, -1);

        int p = 1;
        for (int i=2; i<=n; ++i)
        {
            if (f[i] < f[p])    p = i;
        }

        printf ("%d %I64d\n", p, f[p]);
    }

    return 0;
}