hdu 3836 Equivalent Sets（强连通分量--加边）

Equivalent Sets

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

Total Submission(s): 2798    Accepted Submission(s): 962

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.

 

求将原图的强连通分量缩点，得到有向无环图，求至少加多少条边能够使这个图变成一幅强连通图，max（入度为0的点，出度为0的点）即为答案。

#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
#include"algorithm"
using namespace std;
#define N 20005
#define M 50005
#define min(a,b) (a<b?a:b)
const int inf=1000000;
struct node
{
    int u,v,next;
}e[M];
int t,bcnt,index,stop,ans;
int head[N],dfn[N],low[N],stap[N],mark[N],be[N];
int indeg[N],out[N];
void add(int u,int v)
{
    e[t].u=u;
    e[t].v=v;
    e[t].next=head[u];
    head[u]=t++;
}
void tarjan(int u)
{
    int i,v;
    dfn[u]=low[u]=++index;
    stap[++stop]=u;
    mark[u]=1;
    for(i=head[u];i!=-1;i=e[i].next)
    {
        v=e[i].v;
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(mark[v])
            low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u])
    {
        bcnt++;
        do
        {
            v=stap[stop--];
            mark[v]=0;
            be[v]=bcnt;
        }
        while(u!=v);
        ans++;
    }
}
void solve(int n)
{
    int i;
    memset(dfn,0,sizeof(dfn));
    index=stop=bcnt=0;
    for(i=1;i<=n;i++)
    {
        if(!dfn[i])
            tarjan(i);
    }
}
void work(int n)
{
    int i,j,u,v,t1,t2;
    memset(indeg,0,sizeof(indeg));
    memset(out,0,sizeof(out));
    for(i=1;i<=n;i++)
    {
        u=be[i];
        for(j=head[i];j!=-1;j=e[j].next)
        {
            v=be[e[j].v];
            if(u!=v)
            {
                indeg[v]++;
                out[u]++;
            }
        }
    }
    t1=t2=0;
    for(i=1;i<=bcnt;i++)
    {
        if(indeg[i]==0)
            t1++;
        if(out[i]==0)
            t2++;
    }
    printf("%d\n",t1>t2?t1:t2);
}
int main()
{
    int n,m,u,v;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        t=0;
        memset(head,-1,sizeof(head));
        while(m--)
        {
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        ans=0;
        solve(n);
       // printf("%d\n",ans);
        if(ans==1)
            printf("0\n");
        else
            work(n);
    }
    return 0;
}