题意抽象出来就是求联通块的个数吧,然后添加最少边使图联通。
注意所有人都不会任何语言的时候,答案是n而不是n-1。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<fstream>
#include<sstream>
#include<vector>
#include<string>
#include<cstdio>
#include<bitset>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define MP make_pair
#define eps 1e-8
using namespace std; const int maxn = 111;
int n, m, k, x, fa[maxn];
set<int> lg[maxn];
set<int> :: iterator it; int findset(int x) { return x == fa[x] ? x : fa[x] = findset(fa[x]); } bool check(int i, int j)
{
for(it=lg[i].begin(); it!=lg[i].end(); it++)
if(lg[j].find(*it) != lg[j].end()) return true;
return false;
} int main()
{
scanf("%d%d", &n, &m);
int cnt = 0;
FF(i, 1, n+1)
{
fa[i] = i;
scanf("%d", &k);
if(k == 0) cnt++;
while(k--)
{
scanf("%d", &x);
lg[i].insert(x);
}
}
if(cnt == n)
{
printf("%d\n", n);
return 0;
}
FF(i, 1, n+1) FF(j, i+1, n+1) if(check(i, j))
{
int x = findset(i), y = findset(j);
if(x != y) fa[x] = y;
}
int ans = 0;
FF(i, 1, n+1) if(fa[i] == i) ans++;
printf("%d\n", ans-1);
return 0;
}