hdu 4635 Strongly connected（强联通）

题目链接：hdu 4635 Strongly connected

解题思路

先对给定图做强联通分量，选取出度或者是入度为0的分量中点个数最少的一个，然后其它联通分量算一个，将图分成两部分，做完全图并保证两部分是之间的边均为单向边。

代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <algorithm>

using namespace std;
const int maxn = 100005;
typedef long long ll;

stack<int> S;
vector<int> G[maxn];
int dfsclock, cntscc, sccno[maxn], pre[maxn];

int dfs (int u) {
    int lowu = pre[u] = ++dfsclock;
    S.push(u);

    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (!pre[v]) {
            int lowv = dfs(v);
            lowu = min(lowu, lowv);
        } else if (!sccno[v])
            lowu = min(lowu, pre[v]);
    }

    if (lowu == pre[u]) {
        cntscc++;
        while (true) {
            int x = S.top();
            S.pop();
            sccno[x] = cntscc;
            if (x == u) break;
        }
    }
    return lowu;
}

void findSCC(int n) {
    dfsclock = cntscc = 0;
    memset(pre, 0, sizeof(pre));
    memset(sccno, 0, sizeof(sccno));
    for (int i = 1; i <= n; i++)
        if (!pre[i]) dfs(i);
}

int N, M, C[maxn], in[maxn], ot[maxn];

void init () {
    scanf("%d%d", &N, &M);

    for (int i = 1; i <= N; i++) G[i].clear();

    int u, v;
    for (int i = 0; i < M; i++) {
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
    }

    findSCC(N);
}


ll get(int c1) {
    int c2 = N - c1;
    return 1LL * N * (N-1) - 1LL * c1 * c2 - M;
}

ll solve () {
    if (cntscc == 1) return -1;

    memset(in, 0, sizeof(in));
    memset(ot, 0, sizeof(ot));
    memset(C, 0, sizeof(C));

    for (int i = 1; i <= N; i++) {
        int u = sccno[i];
        C[u]++;

        for (int j = 0; j < G[i].size(); j++) {
            int v = sccno[G[i][j]];
            if (u == v) continue;
            ot[u]++, in[v]++;
        }
    }

    ll ret = 0;
    for (int i = 1; i <= cntscc; i++) {
        if (in[i] && ot[i]) continue;
        ret = max(ret, get(C[i]));
    }

    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        init();
        printf("Case %d: %lld\n", kcas, solve());
    }
    return 0;
}