E - Little W and Contest

https://vjudge.net/contest/386223#problem/E

There are

n

InputThere are several test cases.
The first line contains an integer $T$ OutputFor each test case, output $n$ Sample Input

Sample Output

Sponsor

题意：

　　n个人互不认识，经过一次介绍，两人认识，认识关系具有传递性，n-1次介绍后两联认识；

　　每个人都具有属性1或2，从n个人中选择三个人参加比赛，且至少有两个具有属性2，且三个人互不认识。

　　问每给一次介绍后有多少种凑出来一个队的组合

思路；

利用并查集，将所有的点分成三个部分（三种连通块），每次去不同块中取不满足的情况，

求出总的方案数

merge先计算贡献。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=2e5+10;
const int mod=1e9+7;
ll n = 0;
ll p[100005] = {0};
ll p1[100005] = {0};
ll p2[100005] = {0};
ll In, K;
ll temp;
ll ans;
ll x, y, k = 0;
ll inv(ll a)
{
    return a == 1 ? 1 : (ll)(mod-mod/a)*inv(mod%a)%mod;
}
ll comb(ll n,ll m)
{
    if(m < 0)return 0;
    if(n < m)return 0;
    if(m > n-m) m = n-m;
    ll up = 1;
	ll down = 1;
    for(ll i = 0;i<m;i++)
    {
        up = up * (n-i) % mod;
        down = down * (i+1) % mod;
    }
    return up*inv(down) %mod;
}
ll find(ll x)
{
	if(x != p[x])
		return p[x] = find(p[x]);
	return x;
}

void merge(ll x, ll y)
{
	x = find(x);
	y = find(y);
	p1[y] += p1[x];
	p2[y] += p2[x];
	p[x] = y;
}
int main()
{
	int T = 0;
	cin >> T;
	while(T--)
	{
		cin >> n;
		In = 0; 
		K = 0;
		for(ll i = 1;i<=n;i++) 
		{
			p[i] = i;
			cin >> temp;
			if(temp == 1)
			{
			 	In += 1;
				p1[i] = 1; 
				p2[i] = 0;
			}
			else
			{
				K += 1;
				p1[i] = 0;
				p2[i] = 1;
			}
		}
		ans = (In*comb(K, 2)%mod + comb(K, 3))%mod;
		
		cout << ans <<endl;
		for(int i = 1;i<=n-1;i++)
		{
			k = 0;
			cin >> x >> y;
			x = find(x);
			y = find(y);
			k = (k + p2[x]*p2[y] * (In- p1[x]- p1[y]))%mod;
			k = (k + p2[x]*p2[y] * (In- p2[x]- p2[y]))%mod;
			k = (k + p1[x]*p2[y] * (In- p2[x]- p2[y]))%mod;
			k = (k + p2[x]*p1[y] * (In- p2[x]- p2[y]))%mod;
			merge(x, y);
			ans += mod - k;
			ans %= mod;
			cout << ans << endl;
		}
	}
}

原文地址：https://www.cnblogs.com/SutsuharaYuki/p/13416964.html