Tree and Permutation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 619 Accepted Submission(s): 214
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.
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 ) .
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
大概题意：
给一棵N个节点的树， 求按N个节点全排列的顺序走一遍这棵树所得的路径贡献和。
解题思路：
对于所有按照全排列走过的路径，打表会发现，每两点之间的距离都出现了 (N-1)！次，所以我们只需要求出 所有两点之间的距离之和就可以解得最后的答案了。
也就是说我们只需要求出 ∑dis(i, j) 最后 ∑dis(i, j) * (N-1)! 就是答案了。
∑dis(i, j) 的求法：
我们可以根据当前节点 root 和 它的父节点 fa 的边 v 把一棵树分成两部分， 一部分是 以 fa 为根节点的子树，一部分是以 root 为根节点的子树（即除去这课子树是另一部分）。
如果我们已经预先知道了 fa 到 这棵树每个节点的距离和 sum_dis[ fa ], 以及以 root 为根节点的子树的大小t_size[ root ] ,那么 节点 root 到树上其他节点的距离和我们可以通过 sum_dis[ fa ]转化而来；
因为sum_dis[fa] 减去 v 对上面所分成的树的两部分的影响剩下的就是sum_dis[root] 即sum_dis[root] = sum_dis[fa] - (N-t_size[root])*v - t_size[root]*v ;
因此，我们第一步dfs把 1 到各个节点的距离和求出来，第二步通过对 1 的距离和转换 把其他节点的距离和求出来，最后所有距离和求和得出答案。
AC code：
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#define ll long long int
#define mod 1000000007
#define INF 0x3f3f3f3f
using namespace std; const int MAXN = 1e5+;
struct date{
int v, next, len;
}node[MAXN<<];
ll sum_dis[MAXN];
int t_size[MAXN];
int head[MAXN], cnt;
int N; void init()
{
memset(head, -, sizeof(head));
cnt = ;
memset(sum_dis, , sizeof(sum_dis));
memset(t_size, , sizeof(t_size));
}
void add(int from, int to, int weight)
{
node[++cnt].next = head[from];
head[from] = cnt;
node[cnt].v = to;
node[cnt].len = weight;
}
void dfs(int root, int fa, ll dis)
{
sum_dis[] = (sum_dis[]%mod+dis%mod)%mod;
t_size[root] = ;
for(int x = head[root]; x != -; x =node[x].next)
{
if(node[x].v != fa)
{
dfs(node[x].v, root, (dis+node[x].len)%mod);
t_size[root]+=t_size[node[x].v];
}
}
}
void trans(int root, int fa)
{
int vv;
for(int i = head[root]; i != -; i= node[i].next)
{
vv = node[i].v;
if(node[i].v != fa){
sum_dis[vv] = (sum_dis[root]%mod+((1ll*(N-*t_size[vv])*node[i].len)%mod)%mod)%mod;
if(sum_dis[vv] < ) sum_dis[vv] += mod;
trans(vv, root);
}
}
}
int main()
{
int f, t, w;
while(~scanf("%d", &N))
{
init();
for(int i = ; i < N; i++)
{
scanf("%d%d%d", &f, &t, &w);
add(f, t, w);
add(t, f, w);
}
dfs(, , );
trans(, );
ll P = , ans = ;
for(int i = ; i < N; i++)
P = 1ll*P*i%mod;
for(int i = ; i <= N; i++)
ans = (ans + sum_dis[i])%mod;
ans = ans*P%mod;
printf("%lld\n", ans);
}
return ;
}