'오늘 의 톱 컵' 제1회 호북성 대학 프로 그래 밍 경기 (인터넷 동기 화 경기) C 문제

링크:https://www.nowcoder.com/acm/contest/104/C
우 객 망
Who killed Cock Robin?
I, said the Sparrow, With my bow and arrow,I killed Cock Robin.
Who saw him die?
I, said the Fly.With my little eye,I saw him die.
Who caught his blood?
I, said the Fish,With my little dish,I caught his blood.
Who'll make his shroud?
I, said the Beetle,With my thread and needle,I'll make the shroud.    
.........
All the birds of the air
Fell a-sighing and a-sobbing.
When they heard the bell toll.
For poor Cock Robin.
March 26, 2018
Sparrows are a kind of gregarious animals,sometimes the relationship between them can be represented by a tree.
The Sparrow is for trial, at next bird assizes,we should select a connected subgraph from the whole tree of sparrows as trial objects.
Because the relationship between sparrows is too complex, so we want to leave this problem to you. And your task is to calculate how many different ways can we select a connected subgraph from the whole tree.
입력 설명:

 
  
The first line has a number n to indicate the number of sparrows. 
 
  
The next n-1 row has two numbers x and y per row, which means there is an undirected edge between x and y.
 
  
输出描述:
 
  
 
  
The output is only one integer, the answer module 10000007 (107+7) in a line
 
  

 
  
这道题目想出来状态转移方程后就是一道简单的树形dp
 
  
我们搜每一点后返回的值就相当于以这个点为起点的所有的子联通，难就难在如果父节点
 
  
有多个子节点应该怎么办，解决的办法是遍历每个节点时temp=temp*(dfs(vec[x][i],x)+1);
 
  
为了不搜回去，搜索时保存父节点。
 
  
AC代码 
 
  
#include
#include
#include
#define LL long long
#define maxn 200005
#define mod 10000007
using namespace std;
int n;
vectorvec[maxn];
int x,y;
LL ans;
LL dfs(int x,int pre)
{
    LL temp=1;
    for(int i=0;i