hdu5413 CRB and Roads(topo + bitset)

給定一張有向圖，對於邊e<u,v> ，如果去掉它的話，u仍然可以到達v的話，那麼這條邊就是多餘邊。
顯然的一個思路是倒着做。先topo一遍，然後倒着來，這裏需要用到bitset（也是賽後才知道有這種東西）

const int maxn = 2e4 + 10;
vector<vector<int> > G,mp;
int in[maxn];
int rec[maxn];
int top;
int n, m;
void topo() {
    top = 0;
    for (int i = 1;i <= n;++i)
        if (in[i] == 0) rec[top++] = i;
    for (int i = 0;i < top;++i) {
        int u = rec[i];
        int Size = G[u].size();
        for (int j = 0;j < Size;++j) {
            int v = G[u][j];
            mp[v].push_back(u);
            if (--in[v] == 0) rec[top++] = v;
        }
    }
} 
bitset<maxn> bit[maxn];

int main(int argc, const char * argv[])
{    
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    // clock_t _ = clock();

    int kase;cin >> kase;
    while(kase--) {
        cin >> n >> m;
        G.clear();G.resize(n + 2);
        mp.clear();mp.resize(n + 2);
        int u, v;
        memset(in, 0, sizeof in);
        for (int i = 1;i <= m;++i) {
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            in[v]++;
        }
        topo();
        // for (int i = 0;i < top;++i)
        //     cout << rec[i] << ' ';
        // cout << endl;
        for (int i = 0;i <= n;++i) {
            bit[i].reset(); 
            bit[i].set(i);
        }
        // bit[2] |= bit[1];
        // for (int i = 0;i <= n;++i)
        //     cout << (bit[2][i]?1:0);
        // cout << endl;
        int ans = 0;
        for (int i = 0;i < top;++i) {
            int u = rec[i];
            int Size = mp[u].size();
            for (int j = 0;j < Size;++j) {
                int v = mp[u][j];
                // printf("(%d, %d):", u, v);
                // cout << bit[u] << endl;
                // for (int k = 0;k <= n;++k)
                //     cout << (bit[u][v]?1:0);
                // cout << endl;
                if (bit[u][v]) ans++;
                bit[u] |= bit[v];//dp
            }
            // cout << endl;
        }
        cout << ans << endl;
    }

    // printf("\nTime cost: %.2fs\n", 1.0 * (clock() - _) / CLOCKS_PER_SEC);
    return 0;
}