Strongly connected HDU - 4635

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 \(n\times(n-1)\)，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 \(min\times (n-min)\)
• 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\n", ++cas, (scc_cnt == 1?-1:ans) );
}
int main() {
    int t = 1;scanf("%d", &t);//cin >> t;
    while (t--)solve();
    return 0;
}