Tree and Permutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1421    Accepted Submission(s): 533

 

Problem Description

There are N vertices connected by N−1 edges, each edge has its own length.
The set { 1,2,3,…,N } contains a total of N! unique permutations, let’s say the i-th permutation is Pi and Pi,j is its j-th number.
For the i-th permutation, it can be a traverse sequence of the tree with N vertices, which means we can go from the Pi,1-th vertex to the Pi,2-th vertex by the shortest path, then go to the Pi,3-th vertex ( also by the shortest path ) , and so on. Finally we’ll reach the Pi,N-th vertex, let’s define the total distance of this route as D(Pi) , so please calculate the sum of D(Pi) for all N! permutations.

Input

There are 10 test cases at most.
The first line of each test case contains one integer N ( 1≤N≤105 ) .
For the next N−1 lines, each line contains three integer X, Y and L, which means there is an edge between X-th vertex and Y-th of length L ( 1≤X,Y≤N,1≤L≤109 ) .

Output

For each test case, print the answer module 109+7 in one line.

Sample Input

3 1 2 1 2 3 1 3 1 2 1 1 3 2

Sample Output

16 24

Source

2018中国大学生程序设计竞赛 - 网络选拔赛

看起来很难……事实上就是个纸老虎，(x, y)在遍历序列中相领的情况有(n - 1)!种，对于在(x, y)之间的边都有(n - 1)!的贡献，反过来考虑每条边，对于左侧的点集S和右侧的点集T而言，每条从S到T或从T到S的路径都会造成一次访问，所以每条边的贡献就是

len * Size(S) * Size(T) * 2 * (n - 1)!

一遍dfs就完了

比赛的时候sz数组忘了开long long， wa了两发才过……惨惨

#include <bits/stdc++.h>
#define LL long long
using namespace std;

const int MAXN = 100100;
const int MAXE = 200200;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
const int lim = 100100;

template <typename T> inline void read(T &x) {
    int ch = getchar();
    bool fg = false;
    for (x = 0; !isdigit(ch); ch = getchar()) {
        if (ch == '-') {
            fg = true;
        }
    }
    for (; isdigit(ch); ch = getchar()) {
        x = x * 10 + ch - '0';
    }
    if (fg) {
        x = -x;
    }
}

struct Edge {
    int to, nxt;
    LL len;
    Edge() {}
    Edge(int _to, int _nxt, LL _len) : to(_to), nxt(_nxt), len(_len){}
}E[MAXE];

int h[MAXN], cnt, n;
int fa[MAXN];
LL sz[MAXN];
LL dep[MAXN];
LL fac[MAXN + 10];
LL inv[MAXN + 10];

void init() {
    fac[0] = 1;
    for(int i = 1; i < lim; i++) fac[i] = (long long)fac[i - 1] * i % mod;
    inv[0] = inv[1] = 1;
    for(int i = 2; i < lim; i++) inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod;
    for(int i = 1; i < lim; i++) inv[i] = (long long)inv[i] * inv[i - 1] % mod;
}

typedef pair<int, int> pii;

inline void add_edge(int u, int v, int w) {
    E[++cnt] = Edge(v, h[u], w), h[u] = cnt;
    E[++cnt] = Edge(u, h[v], w), h[v] = cnt;
}

void dfs(int x) {
    sz[x] = 1;
    for(int i = h[x]; i; i = E[i].nxt) {
        int to = E[i].to;
        if(to == fa[x]) continue;
        fa[to] = x;
        dfs(to);
        sz[x] += sz[to];
    }
}

signed main() {
    init();
    while(scanf("%d", &n) != EOF) {
        memset(h, 0, sizeof(h));
        cnt = 0;
        memset(fa, 0, sizeof(fa));
        dep[1] = 0;
        for(int i = 1; i < n; i++) {
            int a, b, c;
            read(a), read(b), read(c);
            add_edge(a, b, c);
        }
        dfs(1);
        LL ans = 0;
        for(int x = 1; x <= n; x++) {
            for(int i = h[x]; i; i = E[i].nxt) {
                int to = E[i].to;
                if(fa[x] == to) continue;
                LL tmp = (n - sz[to]) * sz[to] % mod;
                ans = (ans + ((tmp * 2 % mod) * fac[n - 1] % mod) * (E[i].len % mod)) % mod;
            }
        }
        printf("%lld\n", (ans + mod) % mod);
    }
    return 0;
}