hdu 4635 Strongly connected（强联通）

解题思路

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;
}