正向删点删边太麻烦了怎么办？那就反向加边嘛！

把每个团伙当作点，能联络就代表两者间有边相连，该点蔓延的连通块大小的就是危险程度。

先用vector将边存储。再反向遍历n -> k 并加边 直至当前点(k)加边后的连通块大小大于等于n/2，输出k

#include <bits/stdc++.h>
#define int long long int

using namespace std;

const int N = 2e5 + 7;
int n,m;
int fa[N << 1],tot[N];
vector<int> e[N];

inline int read(){
	int x = 0,f = 1;
	char c = getchar();
	for(;!isdigit(c);c = getchar())
		if(c == '-')f = -1;
	for(;isdigit(c);c = getchar())
		x = (x << 1) + (x << 3) + c - '0';
	return x * f;
}

int get(int x){
	if(fa[x] == x)return x;
	return fa[x] = get(fa[x]);
}

void merge(int x,int y){
	x = get(x);
	y = get(y);
	if(x != y){//注意判是否已经为一家，防止累加
		fa[x] = y;
		tot[y] += tot[x];
	}
}

void print(int x,int y){
	if(get(x) == get(y)) printf("Y\n");
	else printf("N\n");
}

signed main(){
	n = read();
	
	for(int i = 1;i <= n;i ++){
		fa[i] = i;
		tot[i] = 1;
	}
	
	for(int i = 1;i <= n;i ++){
		int k = read();
		while(k --){
			int j = read();
			e[i].push_back(j);
		}
	}
	
	for(int i = n;i > 0;i --){
		for(int j = 0;j < e[i].size();j ++)
			if(e[i][j] > i) merge(e[i][j],i);
		if(tot[get(i)] > (n / 2)){
			printf("%lld",i);
			return 0;
		}
	}
	return 0;
}

类似题目：星球大战(反向加边 + 并查集)