hdu 1827 Summer Holiday (强连通）

/****************************************************

    Summer Holiday(hdu 1827)
    http://acm.hdu.edu.cn/showproblem.php?pid=1827
    强连通
    先通过tarjan找出强连通分量，强连通分量入度为零
    的就一定要打电话，所以入度为零就是最少要打的电
    话数，然后对每个强连通分量找最小值就是最少话费
    
****************************************************/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
using namespace std;

const int mx=1005;
vector<int>g[mx];
stack<int>s;
int dfn[mx],low[mx];
int vs[mx],sccno[mx];
int a[mx],ans[mx],in[mx];
int dfs_cut,scc_cut;

void Init(int n)
{
    dfs_cut=scc_cut=0;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(vs,0,sizeof(vs));
    memset(in,0,sizeof(in));
    memset(ans,0,sizeof(ans));
    memset(sccno,0,sizeof(sccno));
    for (int i=1;i<=n;i++) g[i].clear();
    while (!s.empty()) s.pop();
}


void dfs(int u)
{
    vs[u]=1;
    s.push(u);
    low[u]=dfn[u]=++dfs_cut;
    for (int i=0;i<g[u].size();i++)
    {
        int v=g[u][i];
        if (!vs[v])
        {
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if (!sccno[v]) low[u]=min(low[u],dfn[v]);
    }
    if (low[u]==dfn[u])
    {
        scc_cut++;
        while(1)
        {
            int x=s.top();
            s.pop();
            sccno[x]=scc_cut;
            if (x==u) break;
        }
    }
}


int main()
{
    int n,m,i,j;
    while (~scanf("%d%d",&n,&m))
    {
        Init(n);
        for (i=1;i<=n;i++) scanf("%d",&a[i]);
        int u,v;
        while(m--)
        {
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
        }

        for (i=1;i<=n;i++)
        {
            if (!vs[i]) dfs(i);
        }
        for (i=1;i<=n;i++)
        {
            for (j=0;j<g[i].size();j++)
            {
                int v=g[i][j];
                if (sccno[i]!=sccno[v]) in[sccno[v]]=1;
            }
        }
        for (i=1;i<=n;i++)
        {
            if (!in[sccno[i]])
            {
                if (!ans[sccno[i]]) ans[sccno[i]]=a[i];
                else ans[sccno[i]]=min(ans[sccno[i]],a[i]);
            }
        }
        int ans1=0,cut=0;
        for (i=1;i<=scc_cut;i++)
        {
            ans1+=ans[i];
            if (ans[i]) cut++;
        }
        printf("%d %d
",cut,ans1);
    }
}