UVA 1108 - Mining Your Own Business（双连通分量）

UVA 1108 - Mining Your Own Business

题目链接

题意：给定一个连通图，设置一个些安全点，使得其他任意一些节点崩塌后，其他点都能到一个安全点，问安全点最小数量和情况数

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
, bccno
, dfs_clock, bcc_cnt;
bool iscut
;

vector<int> g
, bcc
;
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 1
while(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;
}