#include <cstdio> #include <cstring> #include <set> using namespace std; /** 在原有的有向连通图中插入尽可能多的边使其不是强连通 由于原图非强连通,因此至少存在一个强连通分支入度或出度为0。而且最后答案中的图中 也应该存在这样的分支。通过添加边无法减少度,因此需要从度为0的分支中找出点数最少的并 使其与外部的点相连且保留原属性 **/ typedef __int64 LL; const int MAXN = 100005; int n, m; int head[MAXN], en; int dfn[MAXN], low[MAXN], cnt; int isin[MAXN], mstk[MAXN], top; int cfz[MAXN], nfz, fz[MAXN]; int id[MAXN], od[MAXN]; struct edge { int v, next; }e[MAXN]; void add(int u, int v) { e[en].v = v; e[en].next = head[u]; head[u] = en++; } void solve(int u) { dfn[u] = low[u] = ++cnt; isin[u] = 1; mstk[++top] = u; for (int i = head[u]; ~i; i = e[i].next) { int v = e[i].v; if (!dfn[v]) { solve(v); low[u] = min(low[v], low[u]); } else if (isin[v]) { low[u] = min(low[u], dfn[v]); } } if (dfn[u] == low[u]) { while (1) { int v = mstk[top]; isin[v] = 0; cfz[nfz]++, --top; fz[v] = nfz; if (u == v) break; } nfz++; } } void minit() { memset(head, -1, sizeof head); en = 0; memset(dfn, 0, sizeof dfn); memset(low, 0, sizeof low); cnt = 0; memset(isin, 0, sizeof isin); top = 0; nfz = 0; memset(cfz, 0, sizeof cfz); memset(id, 0, sizeof id); memset(od, 0, sizeof od); } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); #endif int t, cs = 0; scanf("%d", &t); while (t--) { printf("Case %d: ", ++cs); scanf("%d%d", &n, &m); minit(); for (int i = 0; i< m; ++i) { int u, v; scanf("%d%d", &u, &v); add(u, v); } for (int i= 1; i<= n; ++i) { if (!dfn[i]) solve(i); } if (nfz == 1) { printf("-1\n"); continue; } for (int u = 1; u<= n; ++u) { for (int i = head[u]; ~i; i = e[i].next) { int v = e[i].v; if (fz[u] != fz[v]) { od[fz[u]]++; id[fz[v]]++; } } } LL sum = -1; for (int i = 0; i< nfz; ++i) { LL x = cfz[i], y = n-cfz[i]; if (id[i] == 0 || od[i] == 0) { sum = max(sum, x*(x-1) + y*(y-1) + x*(n-x) - m); } } printf("%I64d\n", sum); } return 0; }