UVA-Proving Equivalences La 4287

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题目很好懂，讲下思路；

把每个命题看成节点，推导视为有向边，得到一个有向图G，那么题意就是如何添加边，使G强连通（每个点都能互相到达），把G中的强连通分量找出后，缩成一个点使G变成一个DAG，接下来设这个DAG上有a个节点出度使0， b个节点入度为0， max（a， b）

这就没什么好证明的了，很显然是这样的。。。

code

#include <stack>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 50005#define next Next#define begin Begin#define mem(a) memset(a, 0, sizeof(a))#define rep(i, j, k) for(int i=j; i<=k; ++i)#define erep(i, u) for(int i=begin[u]; i; i=next[i])void read(int &x){char c=getchar();x=0;while(c<'0' || c>'9')c=getchar();while(c>='0' && c<='9')x=x*10+c-'0', c=getchar();}int begin[N<<1], to[N<<1], next[N<<1], e;void add(int x, int y){to[++e]=y;next[e]=begin[x];begin[x]=e;}int n, m, low[N], pre[N], sccid[N], scc_cnt, clock_;stack<int>s;void dfs(int u){pre[u]=low[u]=++clock_;s.push(u);erep(i, u){int v=to[i];if(!pre[v])dfs(v),low[u]=min(low[u], low[v]);else if(!sccid[v])low[u]=min(low[u], pre[v]);//notice! array low can only reach the point which is in the same SCC;   }if(pre[u] == low[u]){scc_cnt++;while(1){int x=s.top();s.pop();sccid[x]=scc_cnt;if(x == u)break;}}}void solve(){clock_ = scc_cnt = 0;mem(pre);mem(sccid);rep(i, 1, n)if(!pre[i])dfs(i);}int res1[N], res2[N];//no point can reach it, it can't reach any other point;int main(){#ifndef ONLINE_JUDGEfreopen("data.in", "r", stdin);freopen("result.out", "w", stdout);#endifint _;read(_);while(_--){e=0;mem(begin);read(n);read(m);rep(i, 1, m){int u, v;read(u);read(v);add(u, v);}solve();rep(i, 1, scc_cnt)res1[i] = res2[i] = 1;rep(u, 1, n)erep(i, u){int v=to[i];if(sccid[v] != sccid[u])res1[sccid[v]] = res2[sccid[u]] = 0;}int a, b;a = b = 0;rep(i, 1, scc_cnt){if(res1[i]) a++;if(res2[i]) b++;}printf("%d\n", scc_cnt==1? 0: max(a, b));}return 0;}

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