hdu 4635 Strongly connected（强联通）

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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;}

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