原题链接
- 题解：考虑容斥，即假设是有向的完全图，那么边的数量必然是 (n imes(n-1))，考虑并不是强连通图，那么最接近的情况即是两个强连通分量，考虑如何形成，那么就是在原图缩完点的基础上，保留一个入度或者出度为零的
(我就是没想到还有可能是出度为零)强连通分量不与除了自己以外的任何强连通分量连线，那么就是 (min imes (n-min))
- 代码：
#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
const int N = 200100;
const int M = 4000010;
int h[M], ne[M], to[M], idx;
int dfn[N], low[N], times;
int stk[N], instk[N], top;
int scc_cnt, sz[N], id[N];
int d[N],d2[N];
int U[N], V[N];
void add(int u, int v) {ne[idx] = h[u], to[idx] = v, h[u] = idx++;}
void tarjan(int u) {
dfn[u] = low[u] = ++times;
stk[++top] = u;instk[u] = 1;
for (int i = h[u]; ~i ;i = ne[i]) {
int v = to[i];
if (!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if (instk[v]) low[u]=min(low[u], dfn[v]);
}
if (low[u] == dfn[u]) {
scc_cnt++;
while (1) {
int v = stk[top];
instk[v] = 0;
sz[scc_cnt]++;
id[v] = scc_cnt;
top--;
if (v == u) break;
}
}
}
void init() {
memset(h, -1, sizeof h);
idx = scc_cnt = times = top = 0;
memset(dfn, 0, sizeof dfn);
memset(sz, 0, sizeof sz);
memset(d, 0, sizeof d);
memset(d2, 0, sizeof d2);
}
int cas = 0;
void solve() {
ll n, m;scanf("%lld%lld", &n, &m);
init();
for (int i = 1; i <= m; i ++) {
int u, v;
scanf("%d%d", &u, &v);
U[i]= u, V[i] = v;
add(u, v);
}
for (int i = 1; i <= n; i ++) {
if (!dfn[i])tarjan(i);
}
memset(h, -1, sizeof h);
for (int i = 1; i <= m; i ++) {
int uu = id[U[i]];
int vv = id[V[i]];
if (uu == vv)continue;
add(uu, vv);
d[uu] ++;
d2[vv]++;
}
ll ans = 99999999999999;
ll Min1 = ans, Min2 = ans;
for (int i = 1; i <= scc_cnt; i ++) {
if (d[i] == 0) {
Min1 = min(Min1, sz[i]*1ll);
}if (d2[i] == 0) {
Min1 = min(Min1, sz[i] * 1ll);
}
}
ans = n * (n-1) - m - ((n-Min1) * Min1);
printf("Case %d: %lld
", ++cas, (scc_cnt == 1?-1:ans) );
}
int main() {
int t = 1;scanf("%d", &t);//cin >> t;
while (t--)solve();
return 0;
}