1.模板

#include<iostream>
using namespace std;
const int N = 1e4 + 10;
int p[N];
int n, m;

int fd(int x){
	if(x != p[x]){
		p[x] = fd(p[x]);
	}
	return p[x];
}

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++){
		p[i] = i;
	}
	int z, x, y;
	while(m--){
		scanf("%d%d%d", &z, &x, &y);
		int xx = fd(x), yy = fd(y);
		if(z == 1){
			if(xx != yy){
				p[yy] = xx;
			} 
		}else{
			if(xx == yy){
				cout<<"Y"<<endl;
			}else{
				cout<<"N"<<endl;
			}
		}
	}
	return 0;
}

2.亲戚（模板）

#include<iostream>
using namespace std;
const int N = 1e4 + 10;
int p[N];
int n, m, t;

int fd(int x){
	if(x != p[x]){
		p[x] = fd(p[x]);
	}
	return p[x];
}

int main(){
	cin>>n>>m>>t;
	for(int i = 1; i <= n; i++){
		p[i] = i;
	}
	int x, y;
	while(m--){
		cin>>x>>y;
		int xx = fd(x), yy = fd(y);
		if(xx != yy){
			p[yy] = xx;
		}
	}
	while(t--){
		cin>>x>>y;
		int xx = fd(x), yy = fd(y);
		if(xx != yy){
			cout<<"No"<<endl;	
		}else{
			cout<<"Yes"<<endl;
		}
	}
	return 0;
}

3.奶酪（维护连通球距离）

#include<iostream>
#include<cstring>
using namespace std;
const int N = 1e3 + 10;
int p[N], a[N], b[N];
long long x[N], y[N], z[N];
int n, h;
long long r;

long long dis(int i, int j){
	long long sum = 0;
	sum = (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) + (z[i] - z[j]) * (z[i] - z[j]);
	return sum;
}

int fd(int xx){
	if(xx != p[xx]){
		p[xx] = fd(p[xx]);
	}
	return p[xx];
}

int main(){
	int t;
	cin>>t;
	while(t--){
		memset(p, 0, sizeof(p));
		cin>>n>>h>>r;
		for(int i = 1; i <= n; i++){
			p[i] = i;
		}
		int hh = 0, tt = 0;
		for(int i = 1; i <= n; i++){
			cin>>x[i]>>y[i]>>z[i];
			if(z[i] + r >= h){
				a[++hh] = i;
			}
			if(z[i] - r <= 0){
				b[++tt] = i;
			}
			for(int j = 1; j < i; j++){
				if(dis(i, j) <= 4 * r * r){
					int xx = fd(i), yy = fd(j);
					if(xx != yy){
						p[yy] = xx;
					}
				}
			}
		}
		int ok = 0;
		for(int i = 1; i <= hh; i++){
			if(ok) break;
			for(int j = 1; j <= tt; j++){
				if(fd(a[i]) == fd(b[j])){
					ok = 1;
					break;
				}
			}
		}
		if(ok) cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}
	return 0;
}

4.搭配购买（01背包）

可以选择多个集合的云，所以要用 01 背包

#include<iostream>
using namespace std;
const int N = 1e4 + 10;
int p[N];
int c[N], d[N], vv[N], ww[N], f[N];
int n, m, w;

int fd(int x){
	if(x != p[x]){
		p[x] = fd(p[x]);
	}
	return p[x];
}

int main(){
	cin>>n>>m>>w;
	for(int i = 1; i <= n; i++){
		p[i] = i;
	}
	for(int i = 1; i <= n; i++){
		cin>>c[i]>>d[i];
	}
	int u, v;
	for(int i = 1; i <= m; i++){
		cin>>u>>v;
		int xx = fd(u), yy = fd(v);
		if(xx != yy){
			p[yy] = xx;
			c[xx] += c[yy];
			d[xx] += d[yy];
		}
	}
	int cnt = 0;
	for(int i = 1; i <= n; i++){
		if(i == p[i]){
			vv[++cnt] = c[i];
			ww[cnt] = d[i];
		}
	}
	for(int i = 1; i <= cnt; i++){
		for(int j = w; j >= vv[i]; j--){
			f[j] = max(f[j], f[j - vv[i]] + ww[i]);
		}
	}
	cout<<f[w];
	return 0;
}

5.朋友（下标有负数用map）

分别求 -1 和 1 的朋友有多少个（包括他们自己），然后取最小的，就是对数

#include<iostream>
#include<map>
using namespace std;
const int N = 1e4 + 10;
map<int, int> p;
int n, m, t1, t2;

int fd(int x){
	if(x != p[x]){
		p[x] = fd(p[x]);
	}
	return p[x];
}

int main(){
	cin>>n>>m>>t1>>t2;
	for(int i = -m; i <= n; i++){
		p[i] = i;
	}
	int x, y;
	for(int i = 1; i <= t1 + t2; i++){
		cin>>x>>y;
		int xx = fd(x), yy = fd(y);
		if(xx != yy){
			p[yy] = xx;
		}
	}
	int res1 = 0, res2 = 0;
	for(int i = -m; i <= -1; i++){
		int xx = fd(-1), yy = fd(i);
		if(xx == yy){
			res1++;
		}
	}
	for(int i = 1; i <= n; i++){
		int xx = fd(1), yy = fd(i);
		if(xx == yy){
			res2++;
		}
	}
	int res = min(res1, res2);
	cout<<res;
	return 0;
}