[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。

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

  2 

  3 #include <iostream>

  4 #include <cstdio>

  5 #include <algorithm>

  6 #include <cstdlib>

  7 #include <cstring>

  8 #include <map>

  9 #include <queue>

 10 #include <deque>

 11 #include <cmath>

 12 #include <vector>

 13 #include <ctime>

 14 #include <cctype>

 15 #include <set>

 16 #include <bitset>

 17 #include <functional>

 18 #include <numeric>

 19 #include <stdexcept>

 20 #include <utility>

 21 

 22 using namespace std;

 23 

 24 #define mem0(a) memset(a, 0, sizeof(a))

 25 #define mem_1(a) memset(a, -1, sizeof(a))

 26 #define lson l, m, rt << 1

 27 #define rson m + 1, r, rt << 1 | 1

 28 #define define_m int m = (l + r) >> 1

 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)

 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)

 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)

 32 #define rep_down1(a, b) for (int a = b; a > 0; a--)

 33 #define all(a) (a).begin(), (a).end()

 34 #define lowbit(x) ((x) & (-(x)))

 35 #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) {}

 36 #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) {}

 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}

 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}

 39 #define pchr(a) putchar(a)

 40 #define pstr(a) printf("%s", a)

 41 #define sstr(a) scanf("%s", a)

 42 #define sint(a) scanf("%d", &a)

 43 #define sint2(a, b) scanf("%d%d", &a, &b)

 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)

 45 #define pint(a) printf("%d\n", a)

 46 #define test_print1(a) cout << "var1 = " << a << endl

 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl

 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 49 #define mp(a, b) make_pair(a, b)

 50 #define pb(a) push_back(a)

 51 

 52 typedef long long LL;

 53 typedef pair<int, int> pii;

 54 typedef vector<int> vi;

 55 

 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};

 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };

 58 const int maxn = 2e5 + 7;

 59 const int md = 1e9 + 7;

 60 const int inf = 1e9 + 7;

 61 const LL inf_L = 1e18 + 7;

 62 const double pi = acos(-1.0);

 63 const double eps = 1e-6;

 64 

 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}

 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}

 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}

 68 template<class T>T condition(bool f, T a, T b){return f?a:b;}

 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}

 70 int make_id(int x, int y, int n) { return x * n + y; }

 71 

 72 struct Graph {

 73     vector<vector<int> > G;

 74     void clear() { G.clear(); }

 75     void resize(int n) { G.resize(n + 2); }

 76     void add(int u, int v) { G[u].push_back(v); }

 77     vector<int> & operator [] (int u) { return G[u]; }

 78 };

 79 Graph G, pre, suf;

 80 int fa[maxn], f[maxn], ans[maxn], id[maxn], g[maxn];

 81 

 82 void dfs(int n) {

 83     f[n] = 1;

 84     int sz = G[n].size();

 85     rep_up0(i, sz) {

 86         int v = G[n][i];

 87         dfs(v);

 88         f[n] = (LL)f[n] * (1 + f[v]) % md;

 89     }

 90 }

 91 

 92 void getAns(int n) {

 93     int sz = G[n].size();

 94     int fn = fa[n], in = id[n];

 95     pre[n].resize(sz + 2);

 96     suf[n].resize(sz + 2);

 97     pre[n][0] = suf[n][sz + 1] = 1;

 98     if (n == 1) {

 99         ans[n] = f[n];

100         g[n] = 1;

101     }

102     else {

103         g[n] = (1 + (LL)pre[fn][in - 1] % md * suf[fn][in + 1] % md * g[fn]) % md;

104         ans[n] = (LL)f[n] * g[n] % md;

105     }

106     rep_up0(i, sz) {

107         int v = G[n][i];

108         pre[n][i + 1] = (LL)pre[n][i] * (1 + f[v]) % md;

109     }

110     rep_down0(i, sz) {

111         int v = G[n][i];

112         suf[n][i + 1] = (LL)suf[n][i + 2] * (1 + f[v])% md;

113     }

114     rep_up0(i, sz) {

115         int v = G[n][i];

116         getAns(v);

117     }

118 }

119 

120 int main() {

121     //freopen("in.txt", "r", stdin);

122     int n;

123     cin >> n;

124     G.resize(n);

125     pre.resize(n);

126     suf.resize(n);

127     for (int i = 2; i <= n; i ++) {

128         int x;

129         sint(x);

130         G.add(x, i);

131         fa[i] = x;

132         id[i] = G[x].size();

133     }

134     dfs(1);

135     getAns(1);

136     rep_up1(i, n) printf("%d%c", ans[i], i == n? '\n' : ' ');

137     return 0;

138 }

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