LA 4287 Proving Equivalences / 强连通分量

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给你一些命题求最小还需要几次可以证明所有的命题都等价

一个强连通分量里面的题目都是等价的只需缩点后对于DAG图入读为0和出度为0的点两者之中最大值就是答案

如果只有1个强连通分量那么无需证明了

#include <cstdio>#include <cstring>#include <vector>#include <stack>using namespace std;const int maxn = 20010;vector <int> G[maxn];stack <int> s;int pre[maxn];int low[maxn];int sccno[maxn];//每个点所在强联通分量的scc_cnt int in[maxn];//每个强联通分量的入度 int out[maxn];//每个强联通分量的出度int dfs_clock;int scc_cnt;int n, m;void dfs(int u){pre[u] = low[u] = ++dfs_clock;s.push(u);for(int i = 0; i < G[u].size(); i++){int v = G[u][i];if(!pre[v]){dfs(v);low[u] = min(low[u], low[v]);}else if(!sccno[v]){low[u] = min(low[u], pre[v]);}}if(low[u] == pre[u]){scc_cnt++;while(1){int x = s.top();s.pop();sccno[x] = scc_cnt;if(x == u)break;}}}void find_scc(){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(i);}int main(){int T;scanf("%d", &T);while(T--){scanf("%d %d", &n, &m);for(int i = 0; i < n; i++)G[i].clear();for(int i = 0; i < m; i++){int u, v;scanf("%d %d", &u, &v);G[--u].push_back(--v);}find_scc();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;}

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