Strong connected component hdu3836 of a direction graph

Do the problem today. Tarjan algorithm for strongly connected components of graph theory No, so just follow this blog https://www.byvoid.com/blog/scc-tarjan/learn a bit, feel very good
topic is to say equivalence proof, a,b,c, D Mutual proof, the least increase in the number of deduction, a can launch B,b can launch C,c can launch D,d can launch a, at least 4 times. The strong connected components are first identified, then each strong connected component is reduced to a point, and a DAG is obtained. Next, there is a node (remember, here each node relative to the original image of a strong connected component) of the 0,b node of the degree of 0, then max{a,b} is the answer. Note The special case: When the original image is already strong connected, the answer is 0, not 1. The
defines Lowlink (U) as the sequence number (timestamp) of the node U search, and the Pre (U) is the order number of the nodes in the earliest stack that the subtree of U or u can trace back to. By definition,
when Lowlink (u) =pre (u), all nodes on the search subtree with u are a strongly connected component

#include <iostream> #include <vector> #include <stack> #include <stdio.h> #include <string.h

> #define MAXN 50005 using namespace std;
vector<int>g[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;
SCC_CNT is the GCC counter, Sccno[i] is the SCC number for I, stack<int> S; void Dfs (int u) {pre[u]=lowlink[u]=++dfs_clock;//here is used to mark this point processed, the next time Dfs can skip this point s.push (U);//Every search to a point, press into the stack fo
        R (int i=0; i<g[u].size (); i++)//traversing point {int v=g[u][i] connected to u;
            if (!pre[v])//If this point has not been traversed (not on the stack), then search for this point deeply {dfs (v);
        Lowlink[u]=min (Lowlink[u],lowlink[v]);
        else if (!sccno[v])//(on stack) {lowlink[u]=min (lowlink[u],pre[v]);
        } if (Lowlink[u]==pre[u])//found a root {scc_cnt++;
            while (1) {int x=s.top ();//The point in the stack above this point is all out of the stack into a strong connected component S.pop ();
            sccno[x]=scc_cnt;
        if (x==u) break; }} VOID FIND_SCC (int n) {dfs_clock=scc_cnt=0;
    memset (sccno,0,sizeof (SCCNO));
    memset (pre,0,sizeof (pre));
for (int i=0; i<n; i++) if (!pre[i)) Dfs (i);
}//above is template int IN0[MAXN],OUT0[MAXN];
    int main () {int n,m;
        while (~SCANF ("%d%d", &n,&m)) {for (int i=0; i<n; i++) g[i].clear ();
            for (int i=0; i<m; i++) {int u,v;
            scanf ("%d%d", &u,&v);
            u--;
            v--;
        G[u].push_back (v);
        } FIND_SCC (n);
            for (int i=1; i<=scc_cnt; i++) in0[i]=out0[i]=1;//Attach the indentation of each strong connected component to 1 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])//If these two are not a bunch, then there must be a connection between them, then the two pairs of degrees is not 0, because we are asking for 0, so this piece to deal with In0[sccno[v]]=out0
        [sccno[u]]=0;//processed here into 0 for the relative illumination of 1 and is an int a=0,b=0; for (int i=1; i<=scc_cnt;
            i++) {if (In0[i])//1 indicates the need to offset the situation of 0, a++;
        if (Out0[i]) b++;
        int Ans=max (A,B);//MAX offset is the least side if (scc_cnt==1) ans=0 that needs to be added;
    printf ("%d\n", ans);
return 0;
 }