HDU6446 Tree and Permutation（树上DP）

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Tree and Permutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2191    Accepted Submission(s): 826

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≤10^9 ) .

 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

 

 

 

• 题目大意：

         第一行给定n，表示n个顶点，后面n-1行代表边的两个端点和边的权值。

         要输出的是n个顶点全排列对所有情况，按点依次访问的时候，经过的权值和。

         比如说样例1：

         三个顶点，即对1，2，3全排列，共有6种方式。（1->2->3，1->3->2.......）

         按每种情况的访问点的顺序累加权值。

• 思路：

         这题考虑的是每条边对答案的贡献。

         考虑顶点有n个的情况：

         假设顶点x和顶点y有边相连，那么在全排列中这条边会被访问几次？

•  固定x，对其余所有顶点进行全排列，得到(n-1)!种情况。
• 固定y，对其余所有顶点进行全排列，得到(n-1)!种情况。

         显而易见，每条边在全排列中贡献了2*(n-1)！次,再乘上每2个点的最小路径dis[x][y],就是每条边对答案的贡献，即2*(n-1)!*dis[x][y]

         任意两点距离最短路径参考HDU5723 和 HDU2376。

代码：

#include<bits/stdc++.h>
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b)    memset(a, b, sizeof a)
#define REP(i, x, n)    for(int i = x; i <= n; ++i)
#define pi acos(-1.0)
#define Max_N 1001
#define  inf 0x3f3f3f3f      
using namespace std;
const LL mod = 1e9+7;
LL dp[100005];
int sum[100005];
int n;
vector<pair<int,LL> >v[100005];
LL f[100005];
void dfs(int now,int father){
    sum[now] = 1;
    for(int i = 0 ; i < v[now].size() ;  i++){
        int son = v[now][i].first;
        LL  len = v[now][i].second;
        if(son == father)continue;
        dfs(son,now);
        sum[now] += sum[son];
        dp[now]  += (dp[son]+((n-sum[son])%mod*sum[son])%mod*len%mod*2LL*f[n-1]%mod)%mod;
        dp[now] = (dp[now]+mod)%mod;
    }
}
int main()
{
    f[0]=0 ; f[1] = 1;
    for(int i = 2 ; i <= 100000; i ++){
        f[i]=(f[i-1]*i)%mod;
    }
    while(~scanf("%d",&n)){
        memset(dp,0,sizeof(dp));
        memset(sum,0,sizeof(sum));
        for(int i = 0 ; i <= n ; i++)v[i].clear();
        for(int i = 0 ; i < n-1; i ++){
            int x,y;
            LL w;
            scanf("%d %d %lld",&x,&y,&w);
            v[x].push_back(make_pair(y,w));
            v[y].push_back(make_pair(x,w));
        }
        dfs(1,-1);
        printf("%lld
",dp[1]);
    }
}
/*
3
1 2 1
2 3 1
3
1 2 1
1 3 2
*/