hdu2767 Proving Equivalences，有向图强联通，Kosaraju算法

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有向图强联通，Kosaraju算法

缩点后分别入度和出度为0的点的个数 answer = max(a, b);

scc_cnt = 1; answer = 0

#include<cstdio>#include<algorithm>#include<vector>#include<cstring>#include<stack>using namespace std;const int maxn = 20000 + 10;vector<int> G[maxn], G2[maxn];vector<int> S;int vis[maxn], sccno[maxn], scc_cnt;void dfs1(int u){    if(vis[u]) return ;    vis[u] = 1;    for(int i=0; i<G[u].size(); ++i) dfs1(G[u][i]);    S.push_back(u);}void dfs2(int u){    if(sccno[u]) return ;    sccno[u] = scc_cnt;    for(int i=0; i<G2[u].size(); ++i)dfs2(G2[u][i]);}void find_scc(int n){    scc_cnt = 0;    S.clear();    memset(sccno, 0, sizeof sccno );    memset(vis, 0, sizeof vis );    for(int i=0; i<n; ++i) dfs1(i);    for(int i=n-1; i>=0; --i){        if(!sccno[S[i]]) {            scc_cnt++;            dfs2(S[i]);        }    }}int in[maxn], out[maxn];int main(){    int T, n, m;    scanf("%d", &T);    while(T--){        scanf("%d%d", &n, &m);        for(int i=0; i<n; ++i) {            G[i].clear(); G2[i].clear();        }        for(int i=0; i<m; ++i){            int u, v;            scanf("%d%d", &u, &v); u--; v--;            G[u].push_back(v);            G2[v].push_back(u);        }        find_scc(n);        if(scc_cnt==1){            printf("0\n");            continue;        }        memset(in, 0, sizeof in );        memset(out, 0, sizeof out );        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]) {                    out[sccno[u]]++;                    in[sccno[v]]++;                }            }        }        int a = 0, b = 0;        for(int i=1; i<=scc_cnt; ++i){            if(!in[i]) a++;            if(!out[i]) b++;        }        printf("%d\n", max(a, b));    }    return 0;}

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