[tarjan] hdu 3836 Equivalent Sets

主题链接:

http://acm.hdu.edu.cn/showproblem.php?

pid=3836

Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)

Total Submission(s): 2890    Accepted Submission(s): 1006

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

Hint

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

 

Source

field=problem&key=2011+Multi-University+Training+Contest+1+-+Host+by+HNU&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2011 Multi-University Training Contest 1 - Host
by HNU

 

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 | Note
题目意思：

求至少须要增添多少条边，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;
}

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