题面
\(LOJ\)自己找。。
Sol
建立圆方树
考虑枚举起点\(s\)和终点\(t\)
那么答案就是\(s\)到\(t\)间的点双的点数和减去\(s,t\) 设方点权值为点双的点数,圆点的权值为\(-1\) 那么就是求\(s,t\)的路径上的点权和现在考虑中间的点\(x\)
那么它的贡献就是经过它的路径的条数*它的权值 树\(DP\)得解# include# define IL inline# define RG register# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;IL int Input(){ RG int x = 0, z = 1; RG char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1; for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48); return x * z;}const int maxn(4e5 + 5);int n, m, dfn[maxn], low[maxn], idx, sum;int sta[maxn], top, tot, val[maxn], size[maxn];ll ans;struct Edge{ int first[maxn], cnt, nxt[maxn << 1], to[maxn << 1]; IL void Init(){ cnt = 0, Fill(first, -1); } IL void Add(RG int u, RG int v){ nxt[cnt] = first[u], to[cnt] = v, first[u] = cnt++; }} e1, e2;IL void Tarjan(RG int u){ dfn[u] = low[u] = ++idx, sta[++top] = u, size[u] = 1; for(RG int e = e1.first[u]; e != -1; e = e1.nxt[e]){ RG int v = e1.to[e]; if(!dfn[v]){ Tarjan(v), low[u] = min(low[u], low[v]); if(low[v] >= dfn[u]){ RG int x = 0, num = 1; ++tot; do{ x = sta[top--], ++num; e2.Add(tot, x), e2.Add(x, tot); size[tot] += size[x]; } while(x != v); val[tot] = num, size[u] += size[tot]; e2.Add(tot, u), e2.Add(u, tot); } } else low[u] = min(low[u], dfn[v]); }}IL void Dfs(RG int u, RG int ff){ RG int tmp = u <= n; ans += 2LL * size[u] * (sum - size[u]) * val[u]; for(RG int e = e2.first[u]; e != -1; e = e2.nxt[e]){ RG int v = e2.to[e]; if(v != ff){ ans += 2LL * tmp * size[v] * val[u]; tmp += size[v]; Dfs(v, u); } }}int main(){ e1.Init(), e2.Init(); tot = n = Input(), m = Input(); for(RG int i = 1; i <= n; ++i) val[i] = -1; for(RG int i = 1; i <= m; ++i){ RG int u = Input(), v = Input(); e1.Add(u, v), e1.Add(v, u); } for(RG int i = 1; i <= n; ++i) if(!dfn[i]) Tarjan(i), sum = size[i], Dfs(i, 0); printf("%lld\n", ans); return 0;}