zoukankan      html  css  js  c++  java
  • HDU 5647 DZY Loves Connecting 树形dp

    DZY Loves Connecting

    题目连接:

    http://acm.hdu.edu.cn/showproblem.php?pid=5647

    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

    Hint

    题意

    给你一颗树,然后求这棵树的所有连通块大小的和是多少。

    题解:

    树形dp

    dp[i][0]表示以i为根的连通块个数,显然dp[i][0]=PI(dp[v][0]+1),v是i的孩子。

    这个用一个排列组合很容易看出来。

    他的ans实质上是在问,每个点出现在多少个集合的和。

    那么dp[i][1]表示以第i个点为根的所有连通块的点一共出现了多少次。

    现在考虑到v孩子,那么v孩子的父亲x的贡献就增加了:

    (dp[v][0]+1)*dp[x][1]+dp[x][0]*dp[v][1]

    (dp[v][0]+1)*dp[x][1] 表示 从v节点出来的那些节点新增加的贡献。

    dp[x][0]*dp[v][1] 表示 老的贡献又增加了多少。

    然后就完了,作为一个树形dp智障……

    代码

    #pragma comment(linker, "/STACK:102400000,102400000")
    #include<bits/stdc++.h>
    using namespace std;
    const int maxn = 2e5+7;
    const int mod = 1e9+7;
    vector<int>E[maxn];
    long long dp[maxn][2];
    int n;
    void init()
    {
        for(int i=0;i<=n;i++)E[i].clear();
    }
    void dfs(int x,int fa)
    {
        dp[x][0]=dp[x][1]=1;
        for(int i=0;i<E[x].size();i++)
        {
            int v = E[x][i];
            if(v==fa)continue;
            dfs(v,x);
            dp[x][1]=(dp[x][1]*(dp[v][0]+1)+dp[x][0]*dp[v][1])%mod;
            dp[x][0]=(dp[x][0]*(dp[v][0]+1))%mod;
        }
    }
    void solve()
    {
        scanf("%d",&n);
        init();
        for(int i=2;i<=n;i++)
        {
            int x;scanf("%d",&x);
            E[x].push_back(i);
            E[i].push_back(x);
        }
        dfs(1,0);
        long long ans = 0;
        for(int i=1;i<=n;i++)
            ans = (ans+dp[i][1])%mod;
        cout<<ans<<endl;
    }
    int main()
    {
        int t;scanf("%d",&t);
        while(t--)solve();
    }
  • 相关阅读:
    go引入包一直是红色,没有引入的解决办法
    php 把抛出错误记录到日志中
    亚马逊查询接口
    git 合并指定文件到另一个分支
    content-type
    Echarts(饼图Pie)
    DIN 模型速记
    DeepFM 要点速记
    youtube DNN 模型要点速记
    java设计模式之迭代器
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5310614.html
Copyright © 2011-2022 走看看