Ultraviolet A 12167-Proving Equivalences (strongly connected component + point reduction)

Ultraviolet A 12167-Proving Equivalences (strongly connected component + point reduction)
Ultraviolet A 12167-Proving Equivalences

Question Link

Meaning: given some existing proof of equivalence, it is required to be all equal. It must be proved at least several times.

Idea: first obtain the strongly connected component, and then scale down the vertex. On the scaled-down graph, calculate the maximum value of the inbound and outbound degrees of 0 nodes, that is, the number of edges to be added, note that if the entire graph is already strongly connected, it is the answer.

Code:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include using namespace std;const int N = 20005;vector
     
       g[N], scc[N];int pre[N], lowlink[N], sccno[N], dfs_clock, scc_cnt;stack
      
        S;void dfs_scc(int u) {pre[u] = lowlink[u] = ++dfs_clock;S.push(u);for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (!pre[v]) {dfs_scc(v);lowlink[u] = min(lowlink[u], lowlink[v]);} else if (!sccno[v])lowlink[u] = min(lowlink[u], pre[v]);}if (lowlink[u] == pre[u]) {scc_cnt++;while (1) {int x = S.top(); S.pop();sccno[x] = scc_cnt;if (x == u) break;}}}void find_scc(int n) {dfs_clock = scc_cnt = 0;memset(sccno, 0, sizeof(sccno));memset(pre, 0, sizeof(pre));for (int i = 0; i < n; i++)if (!pre[i]) dfs_scc(i);}int in[N], out[N];int T, n, m;int main() {scanf("%d", &T);while (T--) {scanf("%d%d", &n, &m);for (int i = 0; i < n; i++) g[i].clear();int u, v;for (int i = 0; i < m; i++) {scanf("%d%d", &u, &v); u--; v--;g[u].push_back(v);}find_scc(n);for (int i = 1; i <= scc_cnt; i++) in[i] = out[i] = 1;for (int u = 0; u < n; u++)for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (sccno[u] != sccno[v]) in[sccno[v]] = out[sccno[u]] = 0;}int a = 0, b = 0;for (int i = 1; i <= scc_cnt; i++) {if (in[i]) a++;if (out[i]) b++;}int ans = max(a, b);if (scc_cnt == 1) ans = 0;printf("%d\n", ans);}return 0;}