SDUT 算法训练赛E - Reachability from the Capital （tarjan 缩点）

E - Reachability from the Capital

  CodeForces - 999E 

E - Reachability from the Capital

  CodeForces - 999E 

#include<bits/stdc++.h>
using namespace std;
#define maxn 5050
vector<int>G[maxn];
int low[maxn],dfn[maxn],stk[maxn],in[maxn],out[maxn];
int top,cnt,tme,color[maxn],n,m,k;
bool instack[maxn];
void tarjan(int u)
{
    int v;
    low[u]=dfn[u]=++tme;
    stk[top++]=u;
    instack[u]=1;
    for(int i=0; i<G[u].size(); i++)
    {
        v=G[u][i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        if(instack[v])
            low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u])
    {
        cnt++;
        do
        {
            v=stk[--top];
            instack[v]=0;
            color[v]=cnt;
        }
        while(u!=v);
    }
}
void solve()
{
    memset(instack,0,sizeof(instack));
    memset(stk,0,sizeof(stk));
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    int sum1,sum2;
    sum1=sum2=top=cnt=tme=0;
    for(int i=1; i<=n; i++)
        if(!dfn[i])
            tarjan(i);
    for(int i=1; i<=n; i++)
        for(int j=0; j<G[i].size(); j++)
        {
            int nextone=G[i][j];
            if(color[i]!=color[nextone])
            {
                out[color[i]]++;
                in[color[nextone]]++;
            }
        }
    for(int i=1; i<=cnt; i++)
    {
        if(in[i]==0&&color[k]!=i)
        {
            sum1++;
        }
        if(out[i]==0)
        {
            sum2++;
        }
    }
    cout<<sum1<<endl;
}
int main()
{
    int x,y;
    cin>>n>>m>>k;
    while(m--)
    {
        cin>>x>>y;
        G[x].push_back(y);
    }
    solve();
    return 0;
}

构成超级点之后求入度为0且没有 Capital的超级点 计数输出即可