uva 1108 - Mining Your Own Business（强连通）

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题目链接：uva 1108 - Mining Your Own Business

一个联通分量里面如果只有一个割点，则说明该联通分量里面需要一个天井，并且不能建在割点上。特殊情况下为只有一个联通分量的情况，需要选两个点。

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;typedef pair<int,int> pii;typedef long long ll;const int maxn = 1e5 + 5;int N, T[maxn], L[maxn], R[maxn];int cntlock, cntbcc, pre[maxn], low[maxn], bccno[maxn], iscut[maxn];vector<int> G[maxn], BCC[maxn];stack<pii> S;int dfs (int u, int fa) {int lowu = pre[u] = ++cntlock, child = 0;for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];pii e = make_pair(u, v);if (!pre[v]) {child++;S.push(e);int lowv = dfs(v, u);lowu = min(lowu, lowv);if (lowv >= pre[u]) {iscut[u] = 1;BCC[++cntbcc].clear();while (true) {e = S.top(); S.pop();if (bccno[e.first] != cntbcc) {BCC[cntbcc].push_back(e.first);bccno[e.first] = cntbcc;}if (bccno[e.second] != cntbcc) {BCC[cntbcc].push_back(e.second);bccno[e.second] = cntbcc;}if (u == e.first && v == e.second) break;}}} else if (v != fa && pre[v] < pre[u]) {S.push(e);lowu = min(lowu, pre[v]);}}if (fa < 0 && child == 1) iscut[u] = 0;return low[u] = lowu;}void findBCC () {cntlock = cntbcc = 0;memset(pre, 0, sizeof(pre));memset(iscut, 0, sizeof(iscut));memset(bccno, 0, sizeof(bccno));dfs(0, -1);}void init () {int n = 0;for (int i = 1; i <= N; i++) {scanf("%d%d", &L[i], &R[i]);T[n++] = L[i];T[n++] = R[i];}sort(T, T + n);n = unique(T, T + n) - T;for (int i = 0; i < n; i++) G[i].clear();for (int i = 1; i <= N; i++) {int u = lower_bound(T, T + n, L[i]) - T;int v = lower_bound(T, T + n, R[i]) - T;G[u].push_back(v);G[v].push_back(u);}N = n;findBCC();}int main () {int cas = 1;while (scanf("%d", &N) == 1 && N) {init ();ll ans1 = 0, ans2 = 1;if (cntbcc == 1) {ans1 = 2, ans2 = 1LL * N * (N-1) / 2;} else {for (int i = 1; i <= cntbcc; i++) {int c = 0;for (int j = 0; j < BCC[i].size(); j++)if (iscut[BCC[i][j]]) c++;if (c == 1) {ans1++;ans2 *= BCC[i].size() - c;}}}printf("Case %d: %lld %lld\n", cas++, ans1, ans2);}return 0;}

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