算法随笔——图论：无向图三/四元环计数

参考：https://oi-wiki.org/graph/rings-count/

题目链接：P1989 无向图四元环计数

求四元环步骤：

1. 建双向边。
2. 给每条边定向，由度数小的点指向大的，若度数一样则看编号大小。
3. 此时只有这几种情况：

都可以归类为：

4. 枚举起始点 A，枚举 A <--> B（双向边），枚举 B --> C，让 C 点被访问次数 $cnt$ 加 $1$。
5. 四元环数量即为 $cnt[i]\ast (cnt[i]-1)/2$

时间复杂度 $O(n\sqrt{n})$。

具体复杂度证明和三元环求解见 OI WIKI。

代码

// Problem: #191. 无向图四元环计数
// Contest: LibreOJ
// URL: https://loj.ac/p/191
// Memory Limit: 512 MB
// Time Limit: 1000 ms
// Author: Eason
// Date:2024-02-24 18:13:33
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define ll long long
#define INF 0x3f3f3f3f
#define re register
#define il inline
#define gc getchar
inline int read() {
    int f = 1, k = 0;
    char c = getchar();

    while (c < '0' || c > '9') {
        if (c == '-')
            f = -1;

        c = getchar();
    }

    while (c >= '0' && c <= '9')
        k = (k << 1) + (k << 3) + (c ^ 48), c = getchar();

    return k * f;
}
const int N = 2e5 + 5;
int n, m, deg[N];
vector<int> v[N], v1[N];
int cnt[N];
int main() {
    cin >> n >> m;

    for (int i = 1; i <= m; i++) {
        int u =  read(), v = read();
        v1[u].push_back(v);
        v1[v].push_back(u);
        deg[u] ++, deg[v] ++;
    }

    for (int i = 1; i <= n; i++)
        for (int j : v1[i])
            if (deg[i] < deg[j] || (deg[i] == deg[j] && i < j))
                v[i].push_back(j);

    int res = 0;

    for (int i = 1; i <= n; i++) {
        for (int j : v1[i])
            for (int k : v[j])
                if (deg[i] < deg[k] || (deg[i] == deg[k] && i < k))
                    res += cnt[k]++;

        for (int j : v1[i])
            for (int k : v[j])
                cnt[k] = 0;
    }

    cout << res << endl;
    return 0;
}

// Problem: P1989 无向图三元环计数
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P1989
// Memory Limit: 128 MB
// Time Limit: 1000 ms
// Author: Eason
// Date:2024-02-24 18:06:06
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define ll long long
#define INF 0x3f3f3f3f
#define re register
#define il inline
#define gc getchar
inline int read()
{
	int f=1,k=0;
	char c = getchar();
	while (c <'0' || c > '9')
	{
		if (c=='-') f=-1;
		c=getchar();
	}
	while(c >= '0' && c <= '9')  k = (k << 1)+(k << 3)+(c^48),c=getchar();
	return k*f;
}
const int N = 2e5+5;
int n,m,deg[N];
vector<int> v[N],v1[N];
bool vis[N];
int main()
{
	cin >> n >> m;
	for (int i = 1;i <= m;i++)
	{
		int u = read(),v = read();
		v1[u].push_back(v);v1[v].push_back(u);
		deg[u]++,deg[v]++;
	}
	for (int i = 1;i <= n;i++)
		for (auto j : v1[i])
			if (deg[i] < deg[j] || (deg[i] == deg[j] && i < j)) v[i].push_back(j);
	int cnt = 0;
	for (int i = 1;i <= n;i++)
	{
		for (int j :v[i]) vis[j] = 1;
		for (int j:v[i])
			for (int k : v[j]) if (vis[k]) cnt++;
		for (int j :v[i]) vis[j] = 0;
	}
	cout << cnt << endl;	
	return 0;
}