bzoj 4835: 遗忘之树 [树形DP]

4835: 遗忘之树

题意：点分治，选标号最小的重心，上一次重心向下一次重心连有向边，求原树方案数。

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md我真不知道当初比赛时干什么去了...现在一眼秒啊...

\(size[v]=\frac{size[u]}{2}\)时原树只能向编号\(>u\)的点连边，\(O(nlogn)\)

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

typedef long long ll;

const int N = 1e5+5, mo = 1e9+7;

inline int read() {

    char c=getchar(); int x=0,f=1;

    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}

    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}

    return x*f;

}

int n, m, u, v, ind[N], root;

struct edge {int v, ne;} e[N];

int cnt, h[N];

inline void ins(int u, int v) {

	e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;

}

int size[N]; ll f[N];

int find(int u, int p) {

	int ans = u > p;

	for(int i=h[u]; i; i=e[i].ne) ans += find(e[i].v, p);

	return ans;

}

void dp(int u) {

	size[u] = 1; f[u] = 1;

	for(int i=h[u]; i; i=e[i].ne) dp(e[i].v), size[u] += size[e[i].v];

	int half = size[u] & 1 ? 0 : size[u] >> 1;

	for(int i=h[u]; i; i=e[i].ne) {

		int v = e[i].v;

		if(size[v] == half) f[u] = f[u] * f[v] %mo * find(v, u) %mo;

		else f[u] = f[u] * f[v] %mo * size[v] %mo;

	}

}

int main() {

	freopen("in", "r", stdin);

	int T = read();

	while(T--) {

		n = read(); m = read();

		cnt = 0; memset(h, 0, sizeof(h));

		memset(f, 0, sizeof(f));

		memset(ind, 0, sizeof(ind));

		for(int i=1; i<=m; i++) u = read(), v = read(), ins(u, v), ind[v]++;

		for(int i=1; i<=n; i++) if(!ind[i]) {root = i; break;}

		dp(root);

		printf("%lld\n", f[root]);

	}

}