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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 DescriptionThere 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. InputThere 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 ) . OutputFor each test case, print the answer module 109+7 in one line. Sample Input31 2 12 3 131 2 11 3 2Sample Output1624
- 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+;
LL dp[];
int sum[];
int n;
vector<pair<int,LL> >v[];
LL f[];
void dfs(int now,int father){
sum[now] = ;
for(int i = ; 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-]%mod)%mod;
dp[now] = (dp[now]+mod)%mod;
}
}
int main()
{
f[]= ; f[] = ;
for(int i = ; i <= ; i ++){
f[i]=(f[i-]*i)%mod;
}
while(~scanf("%d",&n)){
memset(dp,,sizeof(dp));
memset(sum,,sizeof(sum));
for(int i = ; i <= n ; i++)v[i].clear();
for(int i = ; i < n-; 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(,-);
printf("%lld\n",dp[]);
}
}
/*
3
1 2 1
2 3 1
3
1 2 1
1 3 2
*/
