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;

}