HDU 3836Equivalent Sets——————Kosaraju求强连通分量

Equivalent Sets

Problem Description
To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.

Input
The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.

Output
For each case, output a single integer: the minimum steps needed.

Sample Input
4 0
3 2
1 2
1 3

Sample Output
4
2

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这道题目和HDU2767差不多，只不过没有T罢了
缩点之后，求入度为0的点的个数，出度为0点的个数，取最大值

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5+7;
int cmp[MAXN], ind[MAXN], outd[MAXN], V;
bool used[MAXN];
vector<int> G[MAXN];
vector<int> rG[MAXN];
vector<int> vs;

void add_edge(int x,int y)
{
        G[x].push_back(y);
        rG[y].push_back(x);
}


void dfs(int v)
{
        used[v] = true;
        for(int i=0;i<(int)G[v].size();++i)
                if(!used[G[v][i]])      dfs(G[v][i]);
        vs.push_back(v);
}

void rdfs(int v, int k)
{
        used[v] = true;
        cmp[v]  = k;
        for(int i=0;i<(int)rG[v].size();++i)
                if(!used[rG[v][i]])     rdfs(rG[v][i], k);
}

int scc()
{
        memset(used, 0, sizeof(used));
        vs.clear();
        for(int v=0;v<V;++v)
                if(!used[v])    dfs(v);
        memset(used, 0, sizeof(used));
        int k = 0;
        for(int i = vs.size() - 1; i >= 0; --i)
                if(!used[vs[i]])        rdfs(vs[i], k++);
        return k;
}

int main()
{
        // freopen("../in.txt","r",stdin);
        // freopen("../out.txt","w",stdout);
        int n, m;
        while(cin>>n>>m)
        {
                V = n;
                for(int i=0;i<=n;++i)
                {
                        G[i].clear();
                        rG[i].clear();
                }
                for(int i = 0; i < m; ++i)
                {
                        int x, y;
                        cin>>x>>y;
                        add_edge(x-1,y-1);
                }
                int N = scc();
                if(N==1)
                {
                        cout<<0<<endl;
                        continue;
                }
                memset(ind, 0, sizeof(ind));
                memset(outd,0,sizeof(outd));
                for(int v = 0; v < V; ++v)
                        for(int i = 0;i < (int)G[v].size(); ++i)
                                if(cmp[v]!=cmp[G[v][i]])
                                {
                                        ind[cmp[G[v][i]]]++;
                                        outd[cmp[v]]++;
                                }
                int num1 = 0;
                int num2 = 0;
                for(int i = 0; i < N; ++i)
                {
                        if(!ind[i])      num1++;
                        if(!outd[i])     num2++;
                }
                cout<<max(num1, num2)<<endl;
        }
        return 0;
}