UVA 1108 – Mining Your Own Business

刘汝佳书上都给出了完整的代码

在这里理一下思路：

由题意知肯定存在一个或者多个双连通分量；

假设某一个双连通分量有割顶。那太平井一定不能打在割顶上。

而是选择割顶之外的随意一个点；

假设没有割顶，则要在该双连通分量上打两个井

至于打井方案。见代码

#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 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;
}