[Codeforces543D]Road Improvement

Problem

刚开始每条边都是坏的，现在要选取一个点使得其他点到这个点的路径上最多只有一条坏路，问至少要修好多少条边

Solution

如果以1为根，那么是个简单的树形DP
设根从u转移到v，那么u的父亲会变成v（f[u]需要删除v的贡献），
u的原来的父亲会变成u的孩子（f[u]需要加上原父亲的贡献），
v会成为u的父亲（f[v]加上u的贡献）。
这样从上向下转移即可。

Notice

注意F值的备份和还原

Code

#include<cmath>
#include<deque>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 200000, mo = INF + 7;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
vector<int> edge[N + 5], Left[N + 5], Right[N + 5];
int ans[N + 5], f[N + 5], fa[N + 5], num = 0;
void dp(int u)
{
	f[u] = 1;
	for(auto v : edge[u])
	{
		fa[v] = u;
		dp(v);
		f[u] = (long long)f[u] * (f[v] + 1) % mo;
	}
	int l = 1, r = 1;
	for(auto i = edge[u].begin(); i != edge[u].end(); i++)
	{
		Left[u].push_back(l);
		l = (ll)l * (f[*i] + 1) % mo;
	}
	for(auto i = edge[u].rbegin(); i != edge[u].rend(); i++)
	{
		Right[u].push_back(r);
		r = (ll)r * (f[*i] + 1) % mo;
	}
	reverse(Right[u].begin(), Right[u].end());
}
void dfs(int u)
{
	ans[u] = f[u];
	for(auto i = 0; i < edge[u].size(); i++)
	{
		int v = edge[u][i], uu = f[u], vv = f[v];
		f[u] = (ll)Left[u][i] * Right[u][i] % mo * (f[fa[u]] + 1) % mo;
		f[v] = (ll)f[v] * (f[u] + 1) % mo;
		dfs(v);
		f[u] = uu, f[v] = vv;
	}
}
int sqz()
{
	int n = read();
	rep(i, 2, n) edge[read()].push_back(i);
	fa[1] = 0;
	f[0] = 0;
	dp(1);
	dfs(1);
	rep(i, 1, n) printf("%d%c", ans[i], i == n ? '
' : ' ');
}