圆方树。
我们建立圆方树之后枚举每个起点和终点。
然后设方点的权值为点双的个数，圆点的权值是-1。
之后可以得到答案就是两点之间的点权和。
之后不难发现对于一个点的贡献可以用一遍( ext{dfs})得到,这道题就做完了。
// luogu-judger-enable-o2
#include <bits/stdc++.h>
using namespace std;
const int maxn = 4e5 + 10;
typedef long long ll;
namespace Fast_Input {
template<class T> inline void read(T& res) {
res = 0; char ch; bool sign = 0;
do { ch = getchar(); sign |= ch == '-'; } while(!isdigit(ch));
while(isdigit(ch)) res = (res << 1) + (res << 3) + (ch & 15),ch = getchar();
(sign) && (res = -res);
return;
}
}
using Fast_Input::read;
int n,m,i,j,k,sta_top,extra,tot,s;
int dfn[maxn],low[maxn],sta[maxn],val[maxn],sz[maxn];
ll ans;
struct Graph {
int head[maxn],nxt[maxn << 1],ver[maxn << 1]; int ecnt;
inline Graph() { ecnt = 0; memset(head,-1,sizeof(head)); }
inline void adde(int u,int v) {
ver[++ecnt] = v; nxt[ecnt] = head[u];
head[u] = ecnt; return;
}
inline void readd(int u,int v) {
adde(u,v); adde(v,u); return;
}
} gold,gnew;
inline void cmin(int& x,int y) {
if(x > y) x = y; return;
}
void tarjan(int u) {
dfn[u] = ++tot; low[u] = ++tot; sta[++sta_top] = u; sz[u] = 1;
for(int i = gold.head[u],v;~i;i = gold.nxt[i]) {
v = gold.ver[i];
if(!dfn[v]) {
tarjan(v); cmin(low[u],low[v]);
if(low[v] >= dfn[u]) {
int t = 0,cnt = 1; extra++;
do {
t = sta[sta_top--]; cnt++;
gnew.readd(extra,t);
sz[extra] += sz[t];
} while(t != v);
val[extra] = cnt; sz[u] += sz[extra];
gnew.readd(extra,u);
}
} else cmin(low[u],dfn[v]);
}
}
void dfs(int u,int fa) {
int x = u <= n; ans += 2ll * sz[u] * (s - sz[u]) * val[u];
for(int i = gnew.head[u],v;~i;i = gnew.nxt[i]) {
v = gnew.ver[i]; if(v == fa) continue;
ans += 2ll * x * sz[v] * val[u];
x += sz[v]; dfs(v,u);
}
return;
}
int main() {
read(n); read(m); extra = n;
for(int i = 1;i <= n;i++) val[i] = -1;
for(int i = 1,u,v;i <= m;i++) {
read(u); read(v);
gold.readd(u,v);
}
for(int i = 1;i <= n;i++) {
if(!dfn[i]) {
tarjan(i); s = sz[i]; dfs(i,-1);
}
}
printf("%lld
",ans);
}