Problem:
Score : 300 points

Problem Statement

There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M.
Road i connects City A_i and City B_i.

Snuke can perform the following operation zero or more times:

Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.

After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).

What is the minimum number of roads he must build to achieve the goal?

Constraints

2 \leq N \leq 100,000

1 \leq M \leq 100,000

1 \leq A_i < B_i \leq N

No two roads connect the same pair of cities.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_M B_M

Output

Print the answer.

Sample Input 1

3 1
1 2

Sample Output 1

1

Initially, there are three cities, and there is a road between City 1 and City 2.

Snuke can achieve the goal by building one new road, for example, between City 1 and City 3.
After that,

We can travel between 1 and 2 directly.

We can travel between 1 and 3 directly.

We can travel between 2 and 3 by following both roads (2 - 1 - 3).

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <unistd.h>

#if __has_include(<local.h>)
//#include <local.h>
#endif

using namespace std;

typedef long long int LL;
typedef long long int LLint;

struct dsu {
  public:
    dsu() : _n(0) {}
    dsu(int n) : _n(n), parent_or_size(n, -1) {}

    int merge(int a, int b) {
        assert(0 <= a && a < _n);
        assert(0 <= b && b < _n);
        int x = leader(a), y = leader(b);
        if (x == y) return x;
        if (-parent_or_size[x] < -parent_or_size[y]) std::swap(x, y);
        parent_or_size[x] += parent_or_size[y];
        parent_or_size[y] = x;
        return x;
    }

    bool same(int a, int b) {
        assert(0 <= a && a < _n);
        assert(0 <= b && b < _n);
        return leader(a) == leader(b);
    }

    int leader(int a) {
        assert(0 <= a && a < _n);
        if (parent_or_size[a] < 0) return a;
        return parent_or_size[a] = leader(parent_or_size[a]);
    }

    int size(int a) {
        assert(0 <= a && a < _n);
        return -parent_or_size[leader(a)];
    }

    std::vector<std::vector<int>> groups() {
        std::vector<int> leader_buf(_n), group_size(_n);
        for (int i = 0; i < _n; i++) {
            leader_buf[i] = leader(i);
            group_size[leader_buf[i]]++;
        }
        std::vector<std::vector<int>> result(_n);
        for (int i = 0; i < _n; i++) {
            result[i].reserve(group_size[i]);
        }
        for (int i = 0; i < _n; i++) {
            result[leader_buf[i]].push_back(i);
        }
        result.erase(
            std::remove_if(result.begin(), result.end(),
                           [&](const std::vector<int>& v) { return v.empty(); }),
            result.end());
        return result;
    }

  private:
    int _n;
    // root node: -1 * component size
    // otherwise: parent
    std::vector<int> parent_or_size;
};

//Faster Program
#define endl "\n"

//Faster Coding
#define ED return
#define GET(a) int a;cin >> a
#define DGET(a) double a;cin >> a
#define SGET(a) string a;cin >> a
#define CGET(a) char a;cin >> a
#define PRINT(a) cout << a << "\n"
#define ANS(a,b) int a;cin >> a;cout << b << "\n"
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define SORT(vec) sort(vec.begin(),vec.end())
#define REV(vec) reverse(vec.begin(),vec.end())
#define ALL(x) (x).begin(),(x).end()

//As Function
#define UP(a,b) ((a+(b-1))/(b))
#define IPOW(x,y) LL(pow(x,y))

//Debug
#ifdef LOCAL_H
#define TEST cout << "OK\n"
#define DEBUG(a) a
#else
#define TEST
#define DEBUG(a)
#endif

#ifdef LOCAL_H
[[maybe_unused]]
#endif
const double PI = M_PI;
#ifdef LOCAL_H
[[maybe_unused]]
#endif
const int INF = int(999999999);
#ifdef LOCAL_H
[[maybe_unused]]
#endif
const LL LL_INF = LL(999999999999999999);
#ifdef LOCAL_H
[[maybe_unused]]
#endif
const long long MOD = int(1000000007);

//Settings
#ifdef LOCAL_H
const bool DO_BENCHMARK = true;
#endif
const int DOUBLE_ACCURACY = 15;
const int RUN_SOLVE = 1;

#ifndef LOCAL_H
const bool DO_BENCHMARK = false;
#endif

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(DOUBLE_ACCURACY);
    for(int do_solve_time = 0;do_solve_time < RUN_SOLVE;do_solve_time++){
        void solve();
        if(DO_BENCHMARK){
            clock_t clock_start = clock();
            solve();
            clock_t clock_end = clock();
            const double time = static_cast<double>(clock_end-clock_start)/CLOCKS_PER_SEC*1000.0;
            cout << "\n------------------------------------\n" << time << "[ms]\n";
        }
        else solve();
    }
}

/*------------------------------↓ここからコード↓----------------------------------*/
void solve(){
    int N, M;
    cin >> N >> M;
    dsu d(N);
    for(int i = 0;i < M;i++){
        int a, b;
        cin >> a >> b;
        d.merge(a-1, b-1);
    }
    vector<int> group (N);
    for(int i = 0;i < N;i++){
        group[i] = d.size(i);
    }
    SORT(group);
    int ans = -1;
    for(int i = 0;i < N;){
        ans++;
        i += group[i];
    }
    cout << ans << endl;
    return;
}