Tree and Permutation dfs hdu 6446
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
Recommend
这题看懂题目之后很傻逼的
求全排列中所有1到任一点的距离之和
按边来考虑 就是求贡献了
n个点 n-1条边 所以每条边有（n-1)!
然后只有当要求的两个点的距离在边的两端的时候才有贡献
由于有从左到右 和从右到左两种方法
所以最后每条边的贡献为 2L(n-sz[v])*sz[v] ；
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <cmath> 5 #include <algorithm> 6 #include <set> 7 #include <iostream> 8 #include <map> 9 #include <stack> 10 #include <string> 11 #include <vector> 12 #define pi acos(-1.0) 13 #define eps 1e-6 14 #define fi first 15 #define se second 16 #define lson l,m,rt<<1 17 #define rson m+1,r,rt<<1|1 18 #define bug printf("******\n") 19 #define mem(a,b) memset(a,b,sizeof(a)) 20 #define fuck(x) cout<<"["<<"x="<<x<<"]"<<endl 21 #define f(a) a*a 22 #define sf(n) scanf("%d", &n) 23 #define sff(a,b) scanf("%d %d", &a, &b) 24 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) 25 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d) 26 #define FIN freopen("DATA.txt","r",stdin) 27 #define gcd(a,b) __gcd(a,b) 28 #define lowbit(x) x&-x 29 #pragma comment (linker,"/STACK:102400000,102400000") 30 using namespace std; 31 typedef long long LL; 32 typedef unsigned long long ULL; 33 const int INF = 0x7fffffff; 34 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL; 35 const int mod = 1e9 + 7; 36 const int maxn = 1e6 + 10; 37 int n, tot, head[maxn]; 38 LL d[maxn], sz[maxn]; 39 struct node { 40 int v, nxt; 41 LL w; 42 } edge[maxn << 2]; 43 void init() { 44 tot = 0; 45 mem(head, -1); 46 } 47 void add(int u, int v, LL w) { 48 edge[tot].v = v; 49 edge[tot].w = w; 50 edge[tot].nxt = head[u]; 51 head[u] = tot++; 52 } 53 void dfs(int u, int fa) { 54 sz[u] = 1,d[u]=0; 55 for (int i = head[u]; ~i ; i = edge[i].nxt) { 56 int v = edge[i].v; 57 if (v == fa) continue; 58 dfs(v, u); 59 sz[u] += sz[v]; 60 d[u] = (d[u] + d[v]) % mod; 61 d[u] = (d[u] + edge[i].w * sz[v] % mod * (n - sz[v]) % mod) % mod; 62 } 63 } 64 int main() { 65 while(~sf(n)) { 66 init(); 67 for (int i = 1 ; i < n ; i++) { 68 int u, v; 69 LL w; 70 scanf("%d%d%lld", &u, &v, &w); 71 add(u, v, w); 72 add(v, u, w); 73 } 74 dfs(1, -1); 75 LL f = 1; 76 for (LL i = 1 ; i <= n - 1 ; i++) f = f * i % mod; 77 LL ans = d[1] * f * 2 % mod; 78 printf("%lld\n", ans); 79 } 80 return 0; 81 }
