hdu5647

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Description

  • DZY has an unrooted tree consisting of n nodes labeled from 1 to n.

  • DZY likes connected sets on the tree. A connected set S is a set of nodes, such that every two nodes u,v in S

  • can be connected by a path on the tree, and the path should only contain nodes from

  • S. Obviously, a set consisting of a single node is also considered a connected set.

  • The size of a connected set is defined by the number of nodes which it contains. DZY wants to know the sum of

  • the sizes of all the connected sets. Can you help him count it?

  • The answer may be large. Please output modulo 109+7.

    Input

  • First line contains t denoting the number of testcases.t testcases follow. In each testcase, first line
  • contains n. In lines 2∼n, ith line contains pi, meaning there is an edge between node i and node pi.
  • (1≤pi≤i−1,2≤i≤n) (n≥1， sum of n in all testcases does not exceed 200000)

    Output

  • Output one line for each testcase, modulo 109+7.

    Sample Input

    2
    1
    5
    1
    2
    2
    3

    Sample Output

    1
    42

    题解

  • ans[i]由两部分组成，这个节点做的贡献dp[u](sum[x]+1) 孩子做的贡献dp[x]sum[u]
  • 具体看代码
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=1e6;
const int mod=1e9+7;
int T,n;
ll dp[N],sum[N],head[N],cnt;
ll ans;
struct node{
    int v,nxt;
}e[N];

void readin(){
    scanf("%d",&n);
    rep(i,2,n){
        int x;
        scanf("%d",&x);
        e[cnt].v=i;
        e[cnt].nxt=head[x];
        head[x]=cnt++;
    }
}

void dfs(int u){
    sum[u]=1;
    dp[u]=1;
    for(int i=head[u];i!=-1;i=e[i].nxt){
        int x=e[i].v;
        dfs(x);
        dp[u]=(dp[u]*(sum[x]+1)%mod+dp[x]*sum[u]%mod)%mod;//新加的点做的贡献+这个孩子原来做的贡献 
        sum[u]=(sum[u]*(sum[x]+1))%mod;
    }
    ans=(ans+dp[u])%mod;
}

int main(){
    //ios::sync_with_stdio(false);
    scanf("%d",&T);
    while(T--){
        ans=0;
        cnt=0;
        memset(head,-1,sizeof(head));
        readin();
        dfs(1);
        //cout<<ans<<endl;
        printf("%lld\n",ans);
    }
    return 0;
}