板子-补充

Pollard_Rho

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
 
typedef long long ll;
const int maxn = 105;
ll x[maxn], ans;
ll factor[maxn];

ll multi(ll a, ll b, ll p)	//快速乘
{
	__int128 tmp1=a,tmp2=b,tmp3=p;
	return tmp1*tmp2%tmp3;
}
ll qpow(ll a, ll b, ll p)
{
	ll ans = 1;
	while(b){
		if(b & 1LL)	ans = multi(ans, a, p);
		a = multi(a, a, p);
		b >>= 1;
	}
	return ans;
}
 
bool MR(ll n)
{
	if(n == 2)	return true;
	int s = 20, i, t = 0;
	ll u = n-1;
	while(!(u&1)){
		t++;
		u >>= 1;
	}
	while(s--){
		ll a = rand()%(n-2)+2;
		x[0] = qpow(a, u, n);
		for(i = 1; i <= t; i++){
			x[i] = multi(x[i-1], x[i-1], n);
			if(x[i] == 1 && x[i-1] != 1 && x[i-1] != n-1)	return false;	
		}
		if(x[t] != 1)	return false;
	}
	return true;	
}
 
ll gcd(ll a, ll b)
{
	if(b == 0)	return a;
	else	return gcd(b, a%b);
}
//ll gcd(ll a,ll b){
//	return __gcd(a,b);
//}

ll Pollard_Rho(ll n, int c)
{
	ll i = 1, k = 2, x = rand()%(n-1)+1, y = x;
	while(1){
		i++;
		x = (multi(x, x, n)+c)%n;
		ll p = gcd((y-x+n)%n, n);
		if(p != 1 && p != n)	return p;
		if(y == x)	return n;
		if(i == k){
			y = x;
			k <<= 1;
		}	
	}
}
 
void find(ll n, int c)
{
	if(n == 1)	return;
	if(MR(n)){
		factor[++factor[0]]=n;
		return;
	}
	ll p = n, k = c;
	while(p >= n){
		p = Pollard_Rho(p, c--);
	}
	find(p, k);
	find(n/p, k);
}
 
int main()
{
	ios::sync_with_stdio(0),cin.tie(0);
	srand(time(0));
	ll t,n;
	cin>>t;
	for(int cs=1;cs<=t;++cs){
		cin>>n;
		factor[0]=0;
		find(n,107);
		if(factor[0]==1){
			cout<<"Prime
";
		}
		else{
			ll tmp=factor[1];
			for(int i=2;i<=factor[0];++i){
				if(factor[i]>tmp)tmp=factor[i];
			}
			cout<<tmp<<'
';
		}
	}
	return 0;
}

最大团

复杂度 (O(3^n)).

最大团点的数量=补图中最大独立集点的数量

图的染色问题中，最少需要的颜色的数量=最大团点的数量

const int maxn = 130;
bool mp[maxn][maxn]; //邻接矩阵
int some[maxn][maxn], none[maxn][maxn], all[maxn][maxn];
int n, m, ans;
void dfs(int d, int an, int sn, int nn)
{
	if(!sn && !nn) ans = max(ans, an);
	int u = some[d][0];
	for(int i = 0; i < sn; ++i)
	{
		int v = some[d][i];
		if(mp[u][v]) continue;
		for(int j = 0; j < an; ++j)
		all[d+1][j] = all[d][j];
		all[d+1][an] = v;
		int tsn = 0, tnn = 0;
		for(int j = 0; j < sn; ++j)
		if(mp[v][some[d][j]])
		some[d+1][tsn++] = some[d][j];
		for(int j = 0; j < nn; ++j)
		if(mp[v][none[d][j]])
		none[d+1][tnn++] = none[d][j];
		dfs(d+1, an+1, tsn, tnn);
		some[d][i] = 0, none[d][nn++] = v;
	}
}
int work()
{
	ans = 0;
	for(int i = 0; i < n; ++i) some[1][i] = i+1;
	dfs(1, 0, n, 0);
	return ans;
}

三元环计数

无向图

时间复杂度 ( ext O(msqrt{m}))
https://blog.csdn.net/a_forever_dream/article/details/101441587

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 300010
#define ll long long

int n,m;
struct edge{int y,next;};
edge e[maxn*4];
int len;
int first[maxn];
void buildroad(int x,int y)
{
	e[++len]=(edge){y,first[x]};
	first[x]=len;
}
struct node{int x,y;};
node edges[maxn*2];
int du[maxn],id[maxn],to[maxn],tot[maxn];
inline ll C(int x){return (ll)x*(x-1)/2ll;}

int main()
{
	while(~scanf("%d %d",&n,&m))
	{
		memset(du,0,sizeof(du));//记录每个点的度的数组
		for(int i=1;i<=m;i++)
		scanf("%d %d",&edges[i].x,&edges[i].y),du[edges[i].x]++,du[edges[i].y]++;
		memset(first,0,sizeof(first));len=0;
		for(int i=1,x,y;i<=m;i++)
		{
			x=edges[i].x,y=edges[i].y;
			if(x>y)swap(x,y);//让x成为编号小的点
			if(du[x]>=du[y])buildroad(x,y);//度大的往小的连有向边
			else buildroad(y,x);
		}
		memset(tot,0,sizeof(tot));//别忘了初始化各种数组
		memset(to,0,sizeof(to));//to[i]表示当前点到点i的边是第几条边
		memset(id,0,sizeof(id));//id表示每个点的标记
		for(int i=1;i<=n;i++)
		{
			int x=i;
			for(int j=first[x];j;j=e[j].next)//枚举能到达的点，给他们大商标机
			id[e[j].y]=x,to[e[j].y]=j;//打标记，记录to
			for(int j=first[x];j;j=e[j].next)//再次遍历所有有标记的点
			{
				for(int k=first[e[j].y];k;k=e[k].next)//遍历有标记的点能到达的点
				if(id[e[k].y]==x)tot[j]++,tot[k]++,tot[to[e[k].y]]++;
				//假如能到达一个有标记的点，那么就找到了一个三元环，给这三条边都打上标记
			}
		}
		ll ans=0;
		for(int i=1;i<=len;i++)//统计答案
		ans+=C(tot[i]);
		printf("%lld
",ans);
	}
}

竞赛图

有向完全图的三元环计数
(C(n,3)-sum_{u} C(in_u,2)) ，(in_u) 表示 (u) 的入度。