HDU 3836 equivalent sets (strong connected component)

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Idea: Strongly connected component template problem, a strong connected component of the indentation into a point, to build a new picture. Then on the new diagram, ask for the degrees and degrees of each point.  Assuming that the 0,b of a vertex has a degree of 0, the answer is Max (A, b). You can think of this, the point of 0 is sure to practice the degree of 0 points.

See the code for details:

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include < cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue># Define MAX (a) > (b)? ( A):(B) #define MIN (a) < (b) ( A):(B)) using namespace Std;typedef long long ll;typedef long double ld;const ld EPS = 1e-9, PI = 3.14159265358979323846264 33832795;const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;//& 0x7fffffffconst int seed = 131;const ll INF64 = ll (1e18); const int MAXN = 20000 + 10;int T,N,M,U,V,PRE[MAXN], LOWLINK[MAXN], SCCNO[MAXN], Dfs_clock, scc_cnt;vector<i    nt> g[maxn];stack<int> s;void dfs (int u) {pre[u] = lowlink[u] = ++dfs_clock;    S.push (U);    int len = G[u].size ();        for (int i = 0; i < len; i++) {int v = g[u][i]; if (!pre[v]) {DFS (V);        Lowlink[u] = min (Lowlink[u], lowlink[v]);        } else if (!sccno[v]) {Lowlink[u] = min (Lowlink[u], pre[v]);        }} if (lowlink[u] = = Pre[u]) {scc_cnt++; for (;;) {int x = S.top ();            S.pop ();            SCCNO[X] = scc_cnt;        if (x = = u) break;    }}}void FIND_SCC (int n) {for (int i = 1; I <= n; i++) {if (!pre[i]) DFS (i);    }}int IN0[MAXN], out0[maxn];void init (int n) {dfs_clock = scc_cnt = 0;        for (int i = 1; I <= n; i++) {g[i].clear ();    Sccno[i] = Pre[i] = 0;        }}int Main () {while (~scanf ("%d%d", &n,&m)) {init (n);            for (int i = 1; I <= m; i++) {scanf ("%d%d", &u,&v);        G[u].push_back (v);        } FIND_SCC (n);        if (scc_cnt = = 1) {printf ("0\n"); continue;}        for (int i = 1; I <= scc_cnt; i++) in0[i] = out0[i] = 1;          for (int u = 1; u <= n; u++) {int len = g[u].size ();  for (int j = 0; J < Len; J + +) {int v = g[u][j];            if (sccno[u]! = Sccno[v]) in0[sccno[v]] = out0[sccno[u]] = 0;        }} int a = 0, b = 0;            for (int i = 1; I <= scc_cnt; i++) {if (in0[i]) a++;        if (Out0[i]) b++;        } int ans = max (A, b);    printf ("%d\n", ans); } return 0;}

HDU 3836 equivalent sets (strong connected component)