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题目链接:

http://acm.hdu.edu.cn/showproblem.php?pid=3836

Equivalent Sets

4 0
3 2
1 2
1 3

4
2
Hint
Case 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.
 

求至少需要增添多少条边，使得该图为强连通图。

解题思路：

tarjan求强连通，然后统计入度为0和出度为0的强连通分量个数，两者的最大值即为答案。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 21000
int low[Maxn],dfn[Maxn],dindex,n;
int sta[Maxn],belong[Maxn],bcnt,ss;
bool iss[Maxn];
int de1[Maxn],de2[Maxn];
vector<vector<int> >myv;

void tarjan(int cur)
{
    //printf(":%d\n",cur);
    //system("pause");
    int ne;

    dfn[cur]=low[cur]=++dindex;
    iss[cur]=true;
    sta[++ss]=cur;

    for(int i=0;i<myv[cur].size();i++)
    {
        ne=myv[cur][i];
        if(!dfn[ne])
        {
            tarjan(ne);
            low[cur]=min(low[cur],low[ne]);
        }
        else if(iss[ne]&&dfn[ne]<low[cur])
            low[cur]=dfn[ne];
    }

    if(dfn[cur]==low[cur])
    {
        bcnt++;
        do
        {
            ne=sta[ss--];
            iss[ne]=false;
            belong[ne]=bcnt;
        }while(ne!=cur);
    }
}

void solve()
{
    int i;
    ss=bcnt=dindex=0;
    memset(dfn,0,sizeof(dfn));
    memset(iss,false,sizeof(iss));
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            tarjan(i);
}

int main()
{
    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   while(~scanf("%d",&n))
   {
       myv.clear();
       myv.resize(n+1);
       int m;
       scanf("%d",&m);
       for(int i=1;i<=m;i++)
       {
           int a,b;
           scanf("%d%d",&a,&b);
           myv[a].push_back(b);
       }
       solve();

       if(bcnt==1)
       {
           printf("1\n0\n");
           continue;
       }
       int ansa=0,ansb=0;

       memset(de1,0,sizeof(de1));
       memset(de2,0,sizeof(de2));

       for(int i=1;i<=n;i++)
       {
           for(int j=0;j<myv[i].size();j++)
           {
               int ne=myv[i][j];
               if(belong[i]!=belong[ne])
               {
                   de1[belong[ne]]++;//Èë¶È
                   de2[belong[i]]++; //³ö¶È
               }

           }
       }
       for(int i=1;i<=bcnt;i++)
       {
           if(!de1[i])
                ansa++;
           if(!de2[i])
                ansb++;
       }

       ansb=max(ansb,ansa);

       printf("%d\n",ansb);
   }
    return 0;
}

[tarjan] hdu 3836 Equivalent Sets

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原文地址：http://blog.csdn.net/cc_again/article/details/39716489