Uvalive5135-mining Your Own Business (BCC)

Question Link

N tunnels are connected by some points, with each tunnel connecting two connection points. There is only one tunnel between any two connection points. The task is to install as few Taiping wells and escape devices as possible in these connection points, so that no matter which connection point collapsed, the workers can escape from other Taiping wells and find the minimum number of installation and solutions.

Idea: in fact, this question is equivalent to applying as few black spots as possible in an undirected graph, so that any point can be deleted and each connected component has at least one Black Point. Because different connected components have only one common point at most, it must be a cut point. It can be found that it is not cost-effective to black the cut point, and it is not cost-effective to black the two black spots in a point-connected component. Therefore, only when the point-to-double connected component has only one cut point need to be painted, and any non-cut point needs to be painted black.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;typedef long long ll;const int MAXN = 50005;struct edge{    edge() {}    edge(int uu, int vv) {        u = uu;         v = vv;    }    int u, v;};vector<int> g[MAXN], bcc[MAXN];stack<edge> s;int pre[MAXN], iscut[MAXN], bccno[MAXN];int n, dfs_clock, bcc_cnt, tmp;int dfs(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(u, v);         if (!pre[v]) {            s.push(e);            child++;            int lowv = dfs(v, u);            lowu = min(lowu, lowv);            if (lowv >= pre[u]) {                iscut[u] = true;                 bcc_cnt++;                 bcc[bcc_cnt].clear();                for (;;) {                    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] = 0;    return lowu;}void find_bcc(int n) {    memset(pre, 0, sizeof(pre));    memset(iscut, 0, sizeof(iscut));    memset(bccno, 0, sizeof(bccno));    dfs_clock = bcc_cnt = 0;    for (int i = 1; i <= n; i++)        if (!pre[i])            dfs(i, -1);}int main() {    int t = 1;    while (scanf("%d", &n) && n) {        int u, v;         for (int i = 0; i < MAXN; i++)            g[i].clear();        tmp = 0;        for (int i = 0; i < n; i++) {            scanf("%d%d", &u, &v);            g[u].push_back(v);            g[v].push_back(u);            tmp = max(max(tmp, u), v);        }         find_bcc(tmp);        ll ans1 = 0, ans2 = 1;        if (bcc_cnt == 1) {            ans1 = 2;             ans2 = bcc[1].size() * (bcc[1].size() - 1) / 2;        }        else {            for (int i = 1; i <= bcc_cnt; i++) {                int cnt_cnt = 0;                 for (int j = 0; j < bcc[i].size(); j++)                     if (iscut[bcc[i][j]])                        cnt_cnt++;                if (cnt_cnt == 1) {                    ans1++;                     ans2 *= (ll)(bcc[i].size() - cnt_cnt);                }            }         }        printf("Case %d: %lld %lld\n", t++, ans1, ans2);    }     return 0;}

Uvalive5135-mining Your Own Business (BCC)