题解 点图

传送门

提示：在区间 \([0, 1]\) 中随机一个实数 \(x\)，那么 \(x=0\) 的概率为 \(0\)。

这基本上把做法明说出来了
发现一个随机点恰好落在一个给定面上的概率也为 0
那么不存在四个随机点共面的情况
所以这实际上是道立体几何
手动枚举四个点中有几个是中点
最后答案是

\[三元环个数*3+四元环个数+\sum\frac{\deg(i)(\deg(i)-1)}{2} \]

• 关于四元环计数：见这里

复杂度 \(O(m\sqrt m)\)，卡常严重

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 500010
#define fir first
#define sec second
#define pb push_back
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m;
pair<int, int> e[N];
const ll mod=998244353;

namespace force{
	ll ans, cnt;
	bool mp[1100][1100];
	void solve() {
		ans=1ll*m*(n+m-3);
		for (int i=1; i<=n; ++i) for (int j=1; j<=n; ++j) mp[i][j]=0;
		for (int i=1; i<=m; ++i) mp[e[i].fir][e[i].sec]=mp[e[i].sec][e[i].fir]=1;
		cnt=0;
		for (int i=1; i<=n; ++i) 
			for (int j=i+1; j<=n; ++j) if (mp[i][j])
				for (int k=j+1; k<=n; ++k) if (mp[i][k]&&mp[j][k])
					++cnt;
		// cout<<"cnt: "<<cnt<<endl;
		ans+=cnt*3;
		cnt=0;
		for (int i=1; i<=n; ++i) 
			for (int j=1; j<=n; ++j) if (j!=i&&mp[i][j])
				for (int k=j+1; k<=n; ++k) if (k!=i&&k!=j&&mp[i][k])
					++cnt;
		ans+=cnt;
		cnt=0;
		for (int i=1; i<=n; ++i) 
			for (int j=1; j<=n; ++j) if (j!=i&&mp[i][j])
				for (int k=1; k<=n; ++k) if (k!=i&&k!=j&&mp[j][k])
					for (int l=1; l<=n; ++l) if (l!=i&&l!=j&&l!=k&&mp[k][l]&&mp[i][l])
						++cnt;
		assert(cnt%8==0);
		cout<<"cnt: "<<cnt/8<<endl;
		ans+=cnt/8;
		printf("%lld\n", ans%mod);
	}
}

namespace task1{
	ll deg[N], tim[N], val[N], ans, cnt;
	vector<int> to[N], out[N];
	void solve() {
		ans=1ll*m*(n+m-3);
		for (int i=1; i<=n; ++i) to[i].clear(), out[i].clear(), tim[i]=deg[i]=0;
		for (int i=1; i<=m; ++i) ++deg[e[i].fir], ++deg[e[i].sec];
		for (int i=1; i<=m; ++i) out[e[i].fir].pb(e[i].sec), out[e[i].sec].pb(e[i].fir);
		for (int i=1; i<=m; ++i)
			if (deg[e[i].fir]>deg[e[i].sec]||(deg[e[i].fir]==deg[e[i].sec]&&e[i].fir<e[i].sec)) to[e[i].fir].pb(e[i].sec);
			else to[e[i].sec].pb(e[i].fir);
		cnt=0;
		for (int i=1; i<=n; ++i) {
			for (auto& v:to[i]) tim[v]=i;
			for (auto& v:to[i])
				for (auto& it:to[v]) 
					if (tim[it]==i) ++cnt;
		}
		// cout<<"cnt: "<<cnt<<endl;
		ans+=cnt*3;
		cnt=0;
		for (int i=1; i<=n; ++i) if (out[i].size()) cnt+=1ll*out[i].size()*(out[i].size()-1)/2;
		ans+=cnt;
		cnt=0;
		for (int i=1; i<=n; ++i) tim[i]=val[i]=0;
		for (int i=1; i<=n; ++i) {
			sort(out[i].begin(), out[i].end(), [](int a, int b){return deg[a]>deg[b]||(deg[a]==deg[b]&&a<b);});
			reverse(out[i].begin(), out[i].end());
		}
		for (int i=1; i<=n; ++i)
			for (auto& v:to[i])
				for (auto& it:out[v])
					if (deg[i]>deg[it]||(deg[i]==deg[it]&&i<it)) {
						if (tim[it]!=i) tim[it]=i, val[it]=1;
						else cnt+=val[it]++;
					}
					else break;
		ans+=cnt;
		// cout<<"cnt: "<<cnt<<endl;
		printf("%lld\n", ans%mod);
	}
}

signed main()
{
	freopen("graph.in", "r", stdin);
	freopen("graph.out", "w", stdout);

	int T=read();
	while (T--) {
		n=read(); m=read();
		for (int i=1,u,v; i<=m; ++i) {
			u=read(); v=read();
			e[i]={u, v};
		}
		// force::solve();
		task1::solve();
	}

	return 0;
}