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
");
            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
", max(a, b));
    }
    return 0;
}