hdu 5647 DZY Loves Connecting (树形dp)

DZY Loves Connecting

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 555    Accepted Submission(s): 195

Problem 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 fromS. 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

2151223

 

Sample Output

142
Hint
In the second sample, the 4 edges are (1,2),(2,3),(2,4),(3,5). All the connected sets are {1},{2},{3},{4},{5},{1,2},{2,3},{2,4},{3,5},{1,2,3},{1,2,4},{2,3,4},{2,3,5},{1,2,3,4},{1,2,3,5},{2,3,4,5},{1,2,3,4,5}.If you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++.
 

 

Source

BestCoder Round #76 (div.2)

 

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用f[i]表示包含i的连通块数量，g[i]表示以i为根节点的连通块的数量。

下面给出DP方程：

f[n]=f[n]*(g[son[n][i]]+1)%mod+g[n]*f[son[n][i]]%mod;

f[n]%=mod;

g[n]=g[n]*(g[son[n][i]]+1)%mod;

f[n]*(g[son[n][i]]+1)：对于结点包含n的连通块数量f[n]，当检测到新子结点时，自身大小要乘上以子结点为根的连通块数量，同时还要加上自身

g[n]*f[son[n][i]]：子结点的连通块同样需要更新

g[n]=g[n]*(g[son[n][i]]+1):此处是对以n为根节点的连通块数量进行更新，每多一个子结点，之前的结果都要对该子树进行更新。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<cstdlib>#include<stack>#include <queue>#include <set>#include <map>using namespace std;const int inf = 2147483647;const double PI = acos(-1.0);const double e = 2.718281828459;const int mod = 1000000007;typedef long long LL;#pragma comment(linker, "/STACK:102400000,102400000")//freopen("in.txt","r",stdin); //输入重定向,输入数据将从in.txt文件中读取//freopen("out.txt","w",stdout); //输出重定向,输出数据将保存在out.txt文件中cinconst int nmax = 200005;LL f[nmax], g[nmax];vector<int>son[nmax];int t;void dp_dfs(int x, int father){for (int i = 0; i < son[x].size(); ++i){if (son[x][i] != father){dp_dfs(son[x][i], x);f[x] = (f[x] * g[son[x][i]] % mod + f[x] + g[x] * f[son[x][i]] % mod) % mod;g[x] = g[x] * (g[son[x][i]] + 1) % mod;}}}int main(){int  i, j;int n;scanf("%d", &n);while (n--){scanf("%d", &t);fill(f, f + t + 1, 1);fill(g, g + t + 1, 1);//cout << f[1] << endl;for (i = 2; i <= t; ++i){int x;scanf("%d", &x);son[x].push_back(i);son[i].push_back(x);}dp_dfs(1, -1);LL ans = 0;for (i = 1; i <= t; ++i){ans += f[i];ans %= mod;son[i].clear();}cout << ans << endl;}}