UVA 1108 - Mining Your Own Business（雙連通分量）

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UVA 1108 - Mining Your Own Business

題目連結

題意：給定一個連通圖，設定一個些安全點，使得其他任意一些節點崩塌後，其他點都能到一個安全點，問安全點最小數量和情況數

思路：

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <map>using namespace std;const int N = 50005;struct Edge {int u, v;Edge() {}Edge(int u, int v) {this->u = u;this->v = v;}};int pre[N], bccno[N], dfs_clock, bcc_cnt;bool iscut[N];vector<int> g[N], bcc[N];stack<Edge> S;int dfs_bcc(int u, int fa) {int lowu = pre[u] = ++dfs_clock;int child = 0;for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];Edge e = Edge(u, v);if (!pre[v]) {S.push(e);child++;int lowv = dfs_bcc(v, u);lowu = min(lowu, lowv);if (lowv >= pre[u]) {iscut[u] = true;bcc_cnt++; bcc[bcc_cnt].clear(); //start from 1while(1) {Edge x = S.top(); S.pop();if (bccno[x.u] != bcc_cnt) {bcc[bcc_cnt].push_back(x.u); bccno[x.u] = bcc_cnt;}if (bccno[x.v] != bcc_cnt) {bcc[bcc_cnt].push_back(x.v); bccno[x.v] = bcc_cnt;}if (x.u == u && x.v == v) break;}}} else if (pre[v] < pre[u] && v != fa) {S.push(e);lowu = min(lowu, pre[v]);}}if (fa < 0 && child == 1) iscut[u] = false;return lowu;}int st;void find_bcc() {memset(pre, 0, sizeof(pre));memset(iscut, 0, sizeof(iscut));memset(bccno, 0, sizeof(bccno));dfs_clock = bcc_cnt = 0;dfs_bcc(0, -1);}int n, m;typedef long long ll;void solve() {ll ans1 = 0, ans2 = 1;for (int i = 1; i <= bcc_cnt; i++) {int cut_cnt = 0;for (int j = 0; j < bcc[i].size(); j++)if (iscut[bcc[i][j]]) cut_cnt++;if (cut_cnt == 1) {ans1++; ans2 *= (ll)(bcc[i].size() - cut_cnt);}}if (bcc_cnt == 1) {ans1 = 2; ans2 = (ll)bcc[1].size() * (bcc[1].size() - 1) / 2;}printf(" %lld %lld\n", ans1, ans2);}int main() {int cas = 0;while (~scanf("%d", &m) && m) {int u, v, Max = 0;while (m--) {scanf("%d%d", &u, &v);u--; v--;g[u].push_back(v);g[v].push_back(u);Max = max(Max, u);Max = max(Max, v);}find_bcc();printf("Case %d:", ++cas);solve();for (int i = 0; i <= Max; i++)g[i].clear();}return 0;}

UVA 1108 - Mining Your Own Business（雙連通分量）