Gym102012 - 2018徐州 - G - Rikka with Intersections of Paths (树上差分 + 组合数)

题目链接

给出一棵树，然后在树上标记条路径，问有多少种方法在这条路径中选出条，使得这条路径至少有一个公共点。

枚举每个作为路径交的点，假设经过某点的路径数为，其中有条路径的LCA为这个点，那么此处的计数就是。

这相当于减去了经过这个点的父节点的路径在此处计数中的贡献。

维护经过某点的路径数用树上差分，对于和之间的路径，

。

Time ：1216ms

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;

const int maxn = 3e5 + 7;
const ll mod = 1e9 + 7;

int T, n, m, k, a, b, no, pre[maxn][21], que[maxn];
int head[maxn], deep[maxn], tree[maxn], treey[maxn];
ll fac[maxn], inv[maxn];
struct node
{
	int to, nxt;
}e[maxn << 1];

inline int read()
{
    int cnt = 0;
    char ch = getchar();
    while(ch < '0' || ch > '9')
        ch = getchar();
    while ('0' <= ch && ch <= '9')
    {
        cnt = cnt*10 + ch - '0';
        ch = getchar();
    }
    return cnt;
}

void add(int a, int b)
{
	e[no].to = b, e[no].nxt = head[a];
	head[a] = no++;
	e[no].to = a, e[no].nxt = head[b];
	head[b] = no++;
}

ll qpow(ll a, ll x)
{
	ll res = 1;
	while(x)
	{
		if(x & 1) res = (res*a) % mod;
		a = (a*a) % mod;
		x >>= 1;
	}
	return res;
}

void getfac()
{
	fac[1] = inv[1] = 1LL;
	for(ll i = 2;i < maxn;i++)
		fac[i] = (fac[i-1]*i) % mod;
	inv[maxn-1] = qpow(fac[maxn-1], mod - 2);
	for(ll i = maxn-2;i >= 2;i--)
		inv[i] = (inv[i+1]*(i+1)) % mod;
}

ll C(int u, int v)
{
	if(u < v) return 0;
	if(v == 0 || v == u) return 1LL;
	return fac[u]*inv[v]%mod*inv[u-v]%mod;
}

void dfs2(int u, int Pre)
{
	for(int i = head[u];i != -1;i = e[i].nxt)
	{
		int v = e[i].to;
		if(v == Pre) continue;
		dfs2(v, u);
		tree[u] += tree[v];
    }
}

void init_LCA(int s)
{
	int L = 0, R = 0;
	que[R++] = s, pre[s][0] = 0;
	deep[s] = 0;
	while(L < R)
	{
		int u = que[L++];
		for(int i = 1;i <= 20;i++)
			pre[u][i] = pre[pre[u][i-1]][i-1];
		for(int i = head[u];i != -1;i = e[i].nxt)
		{
			int v = e[i].to;
			if(v == pre[u][0]) continue;
			deep[v] = deep[u] + 1;
			pre[v][0] = u;
			que[R++] = v;
		}
	}
}

int LCA(int u, int v)
{
    if(deep[u] > deep[v]) swap(u, v);
    for(int i = 20;i >= 0;i--)
    	if(deep[u] <= deep[v] - (1<= 0;i--)
    	if(pre[u][i] != pre[v][i]) u = pre[u][i], v = pre[v][i];
    return pre[u][0];
}

void init()
{
	no = 0;
	memset(head, -1, sizeof(head));
	memset(tree, 0, sizeof(tree));
	memset(treey, 0, sizeof(treey));
}

int main()
{
	getfac();
	T = read();
	while(T--)
	{
		n = read(), m = read(), k = read();
		init();
		for(int i = 1;i < n;i++)
		{
			a = read(), b = read();
			add(a, b);
		}
		init_LCA(1);
		for(int i = 1;i <= m;i++)
		{
			a = read(), b = read();
			int o = LCA(a, b);
			tree[a]++, tree[b]++;
			treey[o]++, tree[o]--, tree[pre[o][0]]--;
		}
		ll ans = 0;
		dfs2(1, 0);
		for(int i = 1;i <= n;i++)
			ans = (ans + C(tree[i], k) - C(tree[i] - treey[i], k) + mod) % mod;
		printf("%I64d\n", ans);
	}
	return 0;
}