[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\n", 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[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 2e5 + ;
 const int md = 1e9 + ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?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 + ); }
     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] = ;
     int sz = G[n].size();
     rep_up0(i, sz) {
         int v = G[n][i];
         dfs(v);
         f[n] = (LL)f[n] * ( + f[v]) % md;
     }
 }

 void getAns(int n) {
     int sz = G[n].size();
     int fn = fa[n], in = id[n];
     pre[n].resize(sz + );
     suf[n].resize(sz + );
     pre[n][] = suf[n][sz + ] = ;
     if (n == ) {
         ans[n] = f[n];
         g[n] = ;
     }
     else {
         g[n] = ( + (LL)pre[fn][in - ] % md * suf[fn][in + ] % md * g[fn]) % md;
         ans[n] = (LL)f[n] * g[n] % md;
     }
     rep_up0(i, sz) {
         int v = G[n][i];
         pre[n][i + ] = (LL)pre[n][i] * ( + f[v]) % md;
     }
     rep_down0(i, sz) {
         int v = G[n][i];
         suf[n][i + ] = (LL)suf[n][i + ] * ( + 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 = ; i <= n; i ++) {
         int x;
         sint(x);
         G.add(x, i);
         fa[i] = x;
         id[i] = G[x].size();
     }
     dfs();
     getAns();
     rep_up1(i, n) printf("%d%c", ans[i], i == n? '\n' : ' ');
     return ;
 }