其他分享
首页 > 其他分享> > [CF461B] Appleman and Tree - 树形dp

[CF461B] Appleman and Tree - 树形dp

作者:互联网

## [CF461B] Appleman and Tree - 树形dp

### Description

给你一棵树,有的点是黑色的有的点是白色的,把树分成若干个联通块使得每个联通块有且仅有一个黑点,问有多少种分法。

### Solution

树形 dp,转移时整体枚举一下边割不割即可

$f[i]$ 表示处理掉 i 子树,并且无法上传,$h[i]$ 表示处理掉 i 子树,并且可以上传

``` cpp
#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 1e6 + 5;
const int mod = 1e9 + 7;

int qpow(int p, int q)
{
    return (q & 1 ? p : 1) * (q ? qpow(p * p % mod, q / 2) : 1) % mod;
}

int inv(int p)
{
    return qpow(p, mod - 2);
}

int n, a[N], f[N], h[N];
vector<int> g[N];

void dfs(int p)
{
    int mul_fh = 1;
    for (int q : g[p])
    {
        dfs(q);
        (mul_fh *= f[q] + h[q]) %= mod;
    }
    if (a[p] == 0)
        f[p] += mul_fh;
    if (a[p])
        h[p] += mul_fh;
    else
        for (int q : g[p])
            h[p] += h[q] * mul_fh % mod * inv(f[q] + h[q]) % mod;
    f[p] %= mod;
    h[p] %= mod;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 2; i <= n; i++)
    {
        int x;
        cin >> x;
        ++x;
        g[x].push_back(i);
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    dfs(1);
    for (int i = 1; i <= n; i++)
    {
        // cout << "i=" << i << "\t" << f[i] << "\t" << h[i] << endl;
    }
    cout << h[1] % mod << endl;
}

标签:int,Tree,dfs,Appleman,fh,mul,dp,mod
来源: https://www.cnblogs.com/mollnn/p/14396579.html