题目大意:给定一个无根树,给每条边黑白染色,求出每个点为根时,其他点到根的路径上至多有一条黑边的染色方案数,模$1e9+7$。
题解:树形$DP$不难想到,记$f_u$为以$1$为根时,以$u$为根的子树的方案数,$f_u=prodlimits_{vin son_u}(f_v+1)$
换根也很简单。
但是这题卡模数,换根时要求逆元,其中$f_u$可能等于$1e9+6$,加一后变成$0$,无法求逆。可以求前缀积和后缀积转移
卡点:原$dp$写错
C++ Code:
#include <cstdio>
#include <vector>
#include <cctype>
namespace R {
int x, ch;
inline int read() {
ch = getchar();
while (isspace(ch)) ch = getchar();
for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
return x;
}
}
using R::read;
#define maxn 200010
const int mod = 1e9 + 7;
int head[maxn], cnt;
struct Edge {
int to, nxt;
} e[maxn << 1];
inline void add(int a, int b) {
e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}
int n;
int f[maxn], g[maxn], ans[maxn], l[maxn], r[maxn];
void dfs(int u, int fa = 0) {
f[u] = 1;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v != fa) {
dfs(v, u);
f[u] = static_cast<long long> (1 + f[v]) * f[u] % mod;
}
}
}
void dfs1(int u, int fa = 0) {
std::vector<int> S;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v != fa) S.push_back(v);
}
if (S.size()) {
l[*S.begin()] = g[u];
for (std::vector<int>::iterator it = S.begin() + 1; it != S.end(); it++) {
l[*it] = static_cast<long long> (l[*(it - 1)]) * (f[*(it - 1)] + 1) % mod;
}
r[*(S.end() - 1)] = 1;
if (S.begin() + 1 != S.end()) {
for (std::vector<int>::iterator it = S.end() - 2; true; it--) {
r[*it] = static_cast<long long> (r[*(it + 1)]) * (f[*(it + 1)] + 1) % mod;
if (it == S.begin()) break;
}
}
}
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v != fa) {
g[v] = (static_cast<long long> (l[v]) * r[v] + 1) % mod;
ans[v] = static_cast<long long> (g[v]) * f[v] % mod;
dfs1(v, u);
}
}
}
int main() {
n = read();
for (int i = 1, x; i < n; i++) {
x = read();
add(i + 1, x);
add(x, i + 1);
}
dfs(1);
ans[1] = f[1]; g[1] = 1;
dfs1(1);
for (int i = 1; i <= n; i++) {
printf("%d", ans[i]);
putchar(i == n ? '
' : ' ');
}
return 0;
}