　　　　　　久违的写篇博客吧

A. Maximum Square

题目链接：https://codeforces.com/contest/1243/problem/A

题意：

给定n个栅栏，对这n个栅栏进行任意排序，问可形成的最大正方形面积是多少

分析：

水题。

先排个序，然后暴力枚举正方形边长就可以了

#include<bits/stdc++.h>

#define ios std::ios::sync_with_stdio(false)

#define sd(n) scanf("%d",&n)

#define sdd(n,m) scanf("%d%d",&n,&m)

#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)

#define pd(n) printf("%d\n", (n))

#define pdd(n,m) printf("%d %d\n", n, m)

#define pld(n) printf("%lld\n", n)

#define pldd(n,m) printf("%lld %lld\n", n, m)

#define sld(n) scanf("%lld",&n)

#define sldd(n,m) scanf("%lld%lld",&n,&m)

#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)

#define sf(n) scanf("%lf",&n)

#define sff(n,m) scanf("%lf%lf",&n,&m)

#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)

#define rep(i,a,n) for (int i=a;i<=n;i++)

#define per(i,n,a) for (int i=n;i>=a;i--)

#define mm(a,n) memset(a, n, sizeof(a))

#define pb push_back

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define ll long long

#define MOD 1000000007

#define pi 3.14159265358979323

#define lrt rt<<1

#define rrt rt<<1|1

#define lson l, m, lrt

#define rson m+1, r, rrt

#define debug(x)               cout << #x << ": " << x << endl

#define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;

#define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;

#define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;

using namespace std;

const ll INF (0x3f3f3f3f3f3f3f3fll);

const int inf (0x3f3f3f3f);

template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);

for(res=ch-;isdigit(ch=getchar());res=(res<<)+(res<<)+ch - );flag&&(res=-res);}

template<typename T>void Out(T x){if(x<)putchar('-'),x=-x;if(x>)Out(x/);putchar(x%+'');}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

ll lcm(ll a,ll b){return a*b/gcd(a,b);}

ll pow_mod(ll x,ll n,ll mod){ll res=;while(n){if(n&)res=res*x%mod;x=x*x%mod;n>>=;}return res;}

ll fact_pow(ll n,ll p){ll res=;while(n){n/=p;res+=n;}return res;}

ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=,v=a;while(b){if(b&){r+=v;if(r>p)r-=p;}v<<=;if(v>p)v-=p;b>>=;}return r;}

ll quick_pow(ll a,ll b,ll p){ll r=,v=a%p;while(b){if(b&)r=mult(r,v,p);v=mult(v,v,p);b>>=;}return r;}

bool CH(ll a,ll n,ll x,ll t)

{ll r=quick_pow(a,x,n);ll z=r;for(ll i=;i<=t;i++){r=mult(r,r,n);if(r==&&z!=&&z!=n-)return true;z=r;}return r!=;}

bool Miller_Rabin(ll n)

{if(n<)return false;if(n==)return true;if(!(n&))return false;ll x=n-,t=;while(!(x&)){x>>=;t++;}srand(time(NULL));

ll o=;for(ll i=;i<o;i++){ll a=rand()%(n-)+;if(CH(a,n,x,t))return false;}return true;}

int prime[],minprime[];

void euler(int n)

{int c=,i,j;for(i=;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=;j<=c&&i*prime[j]<=n;j++)

{minprime[i*prime[j]]=prime[j];if(i%prime[j]==)break;}}}

const int N = 1e3 + ;

int a[N];

int main(){

    ios;

    int t;

    cin >> t;

    while(t -- )

    {

        int n;

        cin >> n;

        int a[N];

        rep(i ,  , n - )

        {

            cin >> a[i];

        }

        sort(a,a+n);

        int len = ;

        per(i , n -  , )

        {

            if(a[i] > len){

                len ++;

            }

            else

            {

                break;

            }

        }

        cout << len << '\n';

    }

    return ;

}

B1. Character Swap (Easy Version)

题目链接：https://codeforces.com/contest/1243/problem/B1

题意：

给你两个长度为 n 的字符串 S 和 T ，每次操作你可以交换 S 、T上任意两个位置的字符。问是否能只进行一次操作使 S 串等于 T串。能输出 "Yes" ，不能输出 "No"

分析：

遍历字符串，当 S[i] != T[i] 时，判断从i + 1 到 n是否有s[j] == s[i] && t[j] != s[j]，若不存在则说明无法进行一次操作使得两个字符串相等;

若存在则标记cot为0，继续遍历，倘若又一次出现s[i] != t[i]，则输出"No".

#include<bits/stdc++.h>

#define ios std::ios::sync_with_stdio(false)

#define sd(n) scanf("%d",&n)

#define sdd(n,m) scanf("%d%d",&n,&m)

#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)

#define pd(n) printf("%d\n", (n))

#define pdd(n,m) printf("%d %d\n", n, m)

#define pld(n) printf("%lld\n", n)

#define pldd(n,m) printf("%lld %lld\n", n, m)

#define sld(n) scanf("%lld",&n)

#define sldd(n,m) scanf("%lld%lld",&n,&m)

#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)

#define sf(n) scanf("%lf",&n)

#define sff(n,m) scanf("%lf%lf",&n,&m)

#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)

#define rep(i,a,n) for (int i=a;i<=n;i++)

#define per(i,n,a) for (int i=n;i>=a;i--)

#define mm(a,n) memset(a, n, sizeof(a))

#define pb push_back

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define ll long long

#define MOD 1000000007

#define pi 3.14159265358979323

#define lrt rt<<1

#define rrt rt<<1|1

#define lson l, m, lrt

#define rson m+1, r, rrt

#define debug(x)               cout << #x << ": " << x << endl

#define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;

#define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;

#define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;

using namespace std;

const ll INF (0x3f3f3f3f3f3f3f3fll);

const int inf (0x3f3f3f3f);

template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);

for(res=ch-;isdigit(ch=getchar());res=(res<<)+(res<<)+ch - );flag&&(res=-res);}

template<typename T>void Out(T x){if(x<)putchar('-'),x=-x;if(x>)Out(x/);putchar(x%+'');}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

ll lcm(ll a,ll b){return a*b/gcd(a,b);}

ll pow_mod(ll x,ll n,ll mod){ll res=;while(n){if(n&)res=res*x%mod;x=x*x%mod;n>>=;}return res;}

ll fact_pow(ll n,ll p){ll res=;while(n){n/=p;res+=n;}return res;}

ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=,v=a;while(b){if(b&){r+=v;if(r>p)r-=p;}v<<=;if(v>p)v-=p;b>>=;}return r;}

ll quick_pow(ll a,ll b,ll p){ll r=,v=a%p;while(b){if(b&)r=mult(r,v,p);v=mult(v,v,p);b>>=;}return r;}

bool CH(ll a,ll n,ll x,ll t)

{ll r=quick_pow(a,x,n);ll z=r;for(ll i=;i<=t;i++){r=mult(r,r,n);if(r==&&z!=&&z!=n-)return true;z=r;}return r!=;}

bool Miller_Rabin(ll n)

{if(n<)return false;if(n==)return true;if(!(n&))return false;ll x=n-,t=;while(!(x&)){x>>=;t++;}srand(time(NULL));

ll o=;for(ll i=;i<o;i++){ll a=rand()%(n-)+;if(CH(a,n,x,t))return false;}return true;}

int prime[],minprime[];

void euler(int n)

{int c=,i,j;for(i=;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=;j<=c&&i*prime[j]<=n;j++)

{minprime[i*prime[j]]=prime[j];if(i%prime[j]==)break;}}}

const int N = 2e5 + ;

char s[N] , t[N];

int main()

{

    ios;

    int q;

    cin >> q;

    while(q --)

    {

        int n ;

        cin >> n;

        cin >> s +  >> t +  ;

        int flag =  , cot = ;

        rep(i ,  , n)

        {

            if(!flag) break;

            if(s[i] == t[i])

            continue;

            else

            {

                if(cot == )

                {

                    cout << "No\n";

                    flag = ;

                    break;

                }

                int have = ;

                rep(j , i +  , n)

                {

                    if(s[j] == s[i] && t[j] != s[j])

                    {

                        swap(t[i] , s[j]);

                        cot = ;

                        have = ;

                    }

                }

                if(!have)

                {

                    cout << "No\n";

                    flag = ;

                }

            }

        }

        if(flag)

        cout <<"Yes\n";

    }

    return ;

}

B2. Character Swap (Hard Version)

题目链接：https://codeforces.com/contest/1243/problem/B2

题意：

给你两个长度为 n 的字符串 S 和 T ，每次操作你可以交换 S 、T上任意两个位置的字符。

问在 2n 操作次数内能否是 S 等于 T，若可以则输出"Yes"并打印交换方法，若不可以则输出"No"

分析：

其实仔细观察我们会发现只要能够使S串变为T串，那么它的操作次数一定是在2n以内。

然而我们还是要优先选择优质的交换方法。

当 S[i] != T[i] 从后遍历判断是否存在S[j] == S[i] , 若存在则交换 T[i] 和 S[j] , 否则判断是否存在T[j] == S[i] ，若存在则可以先交换S[i] 和 T[i] ，再交换S[i] = T[j]，若不存在则输出"No"

#include<bits/stdc++.h>

#define ios std::ios::sync_with_stdio(false)

#define sd(n) scanf("%d",&n)

#define sdd(n,m) scanf("%d%d",&n,&m)

#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)

#define pd(n) printf("%d\n", (n))

#define pdd(n,m) printf("%d %d\n", n, m)

#define pld(n) printf("%lld\n", n)

#define pldd(n,m) printf("%lld %lld\n", n, m)

#define sld(n) scanf("%lld",&n)

#define sldd(n,m) scanf("%lld%lld",&n,&m)

#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)

#define sf(n) scanf("%lf",&n)

#define sff(n,m) scanf("%lf%lf",&n,&m)

#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)

#define rep(i,a,n) for (int i=a;i<=n;i++)

#define per(i,n,a) for (int i=n;i>=a;i--)

#define mm(a,n) memset(a, n, sizeof(a))

#define pb push_back

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define ll long long

#define MOD 1000000007

#define pi 3.14159265358979323

#define lrt rt<<1

#define rrt rt<<1|1

#define lson l, m, lrt

#define rson m+1, r, rrt

#define debug(x)               cout << #x << ": " << x << endl

#define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;

#define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;

#define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;

using namespace std;

const ll INF (0x3f3f3f3f3f3f3f3fll);

const int inf (0x3f3f3f3f);

template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);

for(res=ch-;isdigit(ch=getchar());res=(res<<)+(res<<)+ch - );flag&&(res=-res);}

template<typename T>void Out(T x){if(x<)putchar('-'),x=-x;if(x>)Out(x/);putchar(x%+'');}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

ll lcm(ll a,ll b){return a*b/gcd(a,b);}

ll pow_mod(ll x,ll n,ll mod){ll res=;while(n){if(n&)res=res*x%mod;x=x*x%mod;n>>=;}return res;}

ll fact_pow(ll n,ll p){ll res=;while(n){n/=p;res+=n;}return res;}

ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=,v=a;while(b){if(b&){r+=v;if(r>p)r-=p;}v<<=;if(v>p)v-=p;b>>=;}return r;}

ll quick_pow(ll a,ll b,ll p){ll r=,v=a%p;while(b){if(b&)r=mult(r,v,p);v=mult(v,v,p);b>>=;}return r;}

bool CH(ll a,ll n,ll x,ll t)

{ll r=quick_pow(a,x,n);ll z=r;for(ll i=;i<=t;i++){r=mult(r,r,n);if(r==&&z!=&&z!=n-)return true;z=r;}return r!=;}

bool Miller_Rabin(ll n)

{if(n<)return false;if(n==)return true;if(!(n&))return false;ll x=n-,t=;while(!(x&)){x>>=;t++;}srand(time(NULL));

ll o=;for(ll i=;i<o;i++){ll a=rand()%(n-)+;if(CH(a,n,x,t))return false;}return true;}

int prime[],minprime[];

void euler(int n)

{int c=,i,j;for(i=;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=;j<=c&&i*prime[j]<=n;j++)

{minprime[i*prime[j]]=prime[j];if(i%prime[j]==)break;}}}

const int N = 2e5 + ;

char s[N] , t[N];

pair<int , int>hehe;

vector<pair<int , int>>vec;

map<char ,int >haha;

int main()

{

    ios;

    int T;

    cin >> T;

    while(T -- )

    {

        haha.clear();

        vec.clear();

        int n;

        cin >> n;

        cin >> s +  >> t +  ;

        rep(i ,  , n)

        haha[s[i]] ++ , haha[t[i]] ++ ;

        int flag = ;

        rep(i ,  , )

        {

            if(haha[i + 'a'] & )

            {

                flag = ;

                break;

            }

        }

        if(flag)

        {

            cout << "No" << '\n';

            continue;

        }

        rep(i ,  , n)

        {

            if(s[i] == t[i])

            continue;

            else

            {

                int falg = ;

                rep(j , i +  , n)

                {

                    if(s[i] == s[j])

                    {

                        hehe.first = j , hehe.second = i;

                        vec.pb(hehe);

                        swap(s[j] , t[i]);

                        falg = ;

                        break;

                    }

                }

                if(!falg)

                {

                    rep(j , i +  , n)

                    {

                        if(s[i] == t[j])

                        {

                            swap(s[i] , t[i]);

                            hehe.fi = i , hehe.se = i;

                            vec.pb(hehe);

                            swap(s[i] , t[j]);

                            hehe.fi = i , hehe.se = j;

                            vec.pb(hehe);

                            break;

                        }

                    }

                }

            }

        }

        int len = vec.size();

        if(len >=  * n)

        {

            cout << "No" << '\n';

            continue;

        }

        cout << "Yes" << '\n';

        cout << vec.size() << '\n';

        rep(i ,  , vec.size() - )

        cout << vec[i].fi << " " << vec[i].se << '\n';

    }

    return ;

}

C. Tile Painting

题目链接：https://codeforces.com/contest/1243/problem/C

题意：

给你 n 个方格，第 i 个位置的倍数的颜色需要和第 i 个位置的颜色相同，问你最多可以用多少种颜色来填充

分析：

找找规律我们会发现它是有循环节的，而循环节就是n / lcm = 所有因子的gcd

 #include<bits/stdc++.h>

#define ios std::ios::sync_with_stdio(false)

#define sd(n) scanf("%d",&n)

#define sdd(n,m) scanf("%d%d",&n,&m)

#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)

#define pd(n) printf("%d\n", (n))

#define pdd(n,m) printf("%d %d\n", n, m)

#define pld(n) printf("%lld\n", n)

#define pldd(n,m) printf("%lld %lld\n", n, m)

#define sld(n) scanf("%lld",&n)

#define sldd(n,m) scanf("%lld%lld",&n,&m)

#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)

#define sf(n) scanf("%lf",&n)

#define sff(n,m) scanf("%lf%lf",&n,&m)

#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)

#define rep(i,a,n) for (int i=a;i<=n;i++)

#define per(i,n,a) for (int i=n;i>=a;i--)

#define mm(a,n) memset(a, n, sizeof(a))

#define pb push_back

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define ll long long

#define MOD 1000000007

#define pi 3.14159265358979323

#define lrt rt<<1

#define rrt rt<<1|1

#define lson l, m, lrt

#define rson m+1, r, rrt

#define debug(x)               cout << #x << ": " << x << endl

#define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;

#define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;

#define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;

using namespace std;

const ll INF (0x3f3f3f3f3f3f3f3fll);

const int inf (0x3f3f3f3f);

template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);

for(res=ch-;isdigit(ch=getchar());res=(res<<)+(res<<)+ch - );flag&&(res=-res);}

template<typename T>void Out(T x){if(x<)putchar('-'),x=-x;if(x>)Out(x/);putchar(x%+'');}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

ll lcm(ll a,ll b){return a*b/gcd(a,b);}

ll pow_mod(ll x,ll n,ll mod){ll res=;while(n){if(n&)res=res*x%mod;x=x*x%mod;n>>=;}return res;}

ll fact_pow(ll n,ll p){ll res=;while(n){n/=p;res+=n;}return res;}

ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=,v=a;while(b){if(b&){r+=v;if(r>p)r-=p;}v<<=;if(v>p)v-=p;b>>=;}return r;}

ll quick_pow(ll a,ll b,ll p){ll r=,v=a%p;while(b){if(b&)r=mult(r,v,p);v=mult(v,v,p);b>>=;}return r;}

bool CH(ll a,ll n,ll x,ll t)

{ll r=quick_pow(a,x,n);ll z=r;for(ll i=;i<=t;i++){r=mult(r,r,n);if(r==&&z!=&&z!=n-)return true;z=r;}return r!=;}

bool Miller_Rabin(ll n)

{if(n<)return false;if(n==)return true;if(!(n&))return false;ll x=n-,t=;while(!(x&)){x>>=;t++;}srand(time(NULL));

ll o=;for(ll i=;i<o;i++){ll a=rand()%(n-)+;if(CH(a,n,x,t))return false;}return true;}

int prime[],minprime[];

void euler(int n)

{int c=,i,j;for(i=;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=;j<=c&&i*prime[j]<=n;j++)

{minprime[i*prime[j]]=prime[j];if(i%prime[j]==)break;}}}

const int N = 2e5 + ;

vector<ll>haha;

void init(ll n)

{

    ll i ;

    for(i =  ; i * i < n ; i ++)

    {

        if(n % i == )

        haha.pb(i) , haha.pb(n / i);

    }

    if(i * i == n)

    haha.pb(i);

}

int main()

{

    ios;

    haha.clear();

    ll n;

    cin >> n;

    init(n);

    haha.pb(n);

    ll ans = haha[];

    rep(i ,  , haha.size() - )

    ans = gcd(ans , haha[i]);

    cout << ans << '\n';

    return ;

}

D. 0-1 MST

题目链接：https://codeforces.com/contest/1243/problem/D

题意：

给你一张包含 n 个点的完全图，其中有m条边的权值为1，其它的为0。现要求你求出最小生成树的权值

分析：

先将 n 个点存入 set<int> S（表示该点还未在任何连通块里），并用vis进行标记

再对每个点开个set<int>G用来存图

遍历每个点，若 vis[i] = 1 则将 i 从集合S中删除并以 i 为起点进行 bfs 寻找连通块，同时连通块个数 cnt ++。

bfs 过程中遍历S , 若 G[i].count(*it) == 1 ，则说明两点没有连通，若为 0 则说明两点有连通。若连通则将其从S中erase、vis标记为1，并加入队列进行下一轮bfs

最后输出 cnt - 1即可

#include<bits/stdc++.h>

#define ios std::ios::sync_with_stdio(false)

#define sd(n) scanf("%d",&n)

#define sdd(n,m) scanf("%d%d",&n,&m)

#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)

#define pd(n) printf("%d\n", (n))

#define pdd(n,m) printf("%d %d\n", n, m)

#define pld(n) printf("%lld\n", n)

#define pldd(n,m) printf("%lld %lld\n", n, m)

#define sld(n) scanf("%lld",&n)

#define sldd(n,m) scanf("%lld%lld",&n,&m)

#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)

#define sf(n) scanf("%lf",&n)

#define sff(n,m) scanf("%lf%lf",&n,&m)

#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)

#define rep(i,a,n) for (int i=a;i<=n;i++)

#define per(i,n,a) for (int i=n;i>=a;i--)

#define mm(a,n) memset(a, n, sizeof(a))

#define pb push_back

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define ll long long

#define MOD 1000000007

#define pi 3.14159265358979323

#define lrt rt<<1

#define rrt rt<<1|1

#define lson l, m, lrt

#define rson m+1, r, rrt

#define debug(x)               cout << #x << ": " << x << endl

#define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;

#define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;

#define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;

using namespace std;

const ll INF (0x3f3f3f3f3f3f3f3fll);

const int inf (0x3f3f3f3f);

template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);

for(res=ch-;isdigit(ch=getchar());res=(res<<)+(res<<)+ch - );flag&&(res=-res);}

template<typename T>void Out(T x){if(x<)putchar('-'),x=-x;if(x>)Out(x/);putchar(x%+'');}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

ll lcm(ll a,ll b){return a*b/gcd(a,b);}

ll pow_mod(ll x,ll n,ll mod){ll res=;while(n){if(n&)res=res*x%mod;x=x*x%mod;n>>=;}return res;}

ll fact_pow(ll n,ll p){ll res=;while(n){n/=p;res+=n;}return res;}

ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=,v=a;while(b){if(b&){r+=v;if(r>p)r-=p;}v<<=;if(v>p)v-=p;b>>=;}return r;}

ll quick_pow(ll a,ll b,ll p){ll r=,v=a%p;while(b){if(b&)r=mult(r,v,p);v=mult(v,v,p);b>>=;}return r;}

bool CH(ll a,ll n,ll x,ll t)

{ll r=quick_pow(a,x,n);ll z=r;for(ll i=;i<=t;i++){r=mult(r,r,n);if(r==&&z!=&&z!=n-)return true;z=r;}return r!=;}

bool Miller_Rabin(ll n)

{if(n<)return false;if(n==)return true;if(!(n&))return false;ll x=n-,t=;while(!(x&)){x>>=;t++;}srand(time(NULL));

ll o=;for(ll i=;i<o;i++){ll a=rand()%(n-)+;if(CH(a,n,x,t))return false;}return true;}

int prime[],minprime[];

void euler(int n)

{int c=,i,j;for(i=;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=;j<=c&&i*prime[j]<=n;j++)

{minprime[i*prime[j]]=prime[j];if(i%prime[j]==)break;}}}

const int N = 2e5 + ;

set<int>G[N];

set<int>s;

set<int>::iterator it;

int n , m;

int vis[N];

void bfs(int x)

{

    queue<int>q;

    q.push(x);

    s.erase(x);

    vis[x] = ;

    while(!q.empty())

    {

        int k = q.front();

        q.pop();

        for(it = s.begin() ; it != s.end() ;)

        {

            int ha = *it ++;

            if(G[k].count(ha) == )

            {

                vis[ha] = ;

                q.push(ha);

                s.erase(ha);

            }

        }

    }

}

int main()

{

    ios;

    cin >> n >> m;

    rep(i ,  , n)

    s.insert(i);

    rep(i ,  , m)

    {

        int x , y;

        cin >> x >> y;

        G[x].insert(y);

        G[y].insert(x);

    }

    int ans = ;

    rep(i ,  , n)

    {

        if(!vis[i])

        {

            ans ++;

            bfs(i);

        }

    }

    cout << ans -  << '\n';

    return ;

}

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Codeforces Round #599 (Div. 2)