[codeforces-543-D div1]树型DP

题意：给一棵树的边标上0或1，求以节点i为源点，其它点到i的唯一路径上的1的边数不超过1条的方案数，输出所有i的答案。

思路：令f[i]表示以节点i为源点，只考虑子树i时的方案数，ans[i]为最后答案，fa[i]为i的父亲，则不难得出以下转移方程：

f[i] = ∏(1 + f[v]),v是i的儿子      ans[i] = f[i] * (1 + ans[fa[i]] / (1 + f[i]))

由于除法取模运算的存在，不得不对1+f[i]求逆元，但1+f[i]可能等于MOD，于是这种情况下结果就是错的了，不能用这个公式求。

令g[i] = ans[fa[i]] / (1 + f[i])，注意到g[i]实际上等于∏(1 + f[v]) * g[fa[i]],v是i的兄弟，于是可以增加一个前缀积数组和一个后缀积数组用来得到∏(1 + f[v])，就不难得到g[i]和最后的答案了。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt5(name, a, b, c, d, e) name(int a = 0, int b = 0, int c = 0, int d = 0, int e = 0): a(a), b(b), c(c), d(d), e(e) {}
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)

typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 2e5 + 7;
const int md = 1e9 + 7;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

struct Graph {
    vector<vector<int> > G;
    void clear() { G.clear(); }
    void resize(int n) { G.resize(n + 2); }
    void add(int u, int v) { G[u].push_back(v); }
    vector<int> & operator [] (int u) { return G[u]; }
};
Graph G, pre, suf;
int fa[maxn], f[maxn], ans[maxn], id[maxn], g[maxn];

void dfs(int n) {
    f[n] = 1;
    int sz = G[n].size();
    rep_up0(i, sz) {
        int v = G[n][i];
        dfs(v);
        f[n] = (LL)f[n] * (1 + f[v]) % md;
    }
}

void getAns(int n) {
    int sz = G[n].size();
    int fn = fa[n], in = id[n];
    pre[n].resize(sz + 2);
    suf[n].resize(sz + 2);
    pre[n][0] = suf[n][sz + 1] = 1;
    if (n == 1) {
        ans[n] = f[n];
        g[n] = 1;
    }
    else {
        g[n] = (1 + (LL)pre[fn][in - 1] % md * suf[fn][in + 1] % md * g[fn]) % md;
        ans[n] = (LL)f[n] * g[n] % md;
    }
    rep_up0(i, sz) {
        int v = G[n][i];
        pre[n][i + 1] = (LL)pre[n][i] * (1 + f[v]) % md;
    }
    rep_down0(i, sz) {
        int v = G[n][i];
        suf[n][i + 1] = (LL)suf[n][i + 2] * (1 + f[v])% md;
    }
    rep_up0(i, sz) {
        int v = G[n][i];
        getAns(v);
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    int n;
    cin >> n;
    G.resize(n);
    pre.resize(n);
    suf.resize(n);
    for (int i = 2; i <= n; i ++) {
        int x;
        sint(x);
        G.add(x, i);
        fa[i] = x;
        id[i] = G[x].size();
    }
    dfs(1);
    getAns(1);
    rep_up1(i, n) printf("%d%c", ans[i], i == n? '
' : ' ');
    return 0;
}