Hawk-and-Chicken

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Kids in kindergarten enjoy playing a game called Hawk-and-Chicken. But there always exists a big problem: every kid in this game want to play the role of Hawk.

So the teacher came up with an idea: Vote. Every child have some nice handkerchiefs, and if he/she think someone is suitable for the role of Hawk, he/she gives a handkerchief to this kid, which means this kid who is given the handkerchief win the support. Note
the support can be transmitted. Kids who get the most supports win in the vote and able to play the role of Hawk.(A note:if A can win

support from B(A != B) A can win only one support from B in any case the number of the supports transmitted from B to A are many. And A can‘t win the support from himself in any case.

If two or more kids own the same number of support from others, we treat all of them as winner.

Here‘s a sample: 3 kids A, B and C, A gives a handkerchief to B, B gives a handkerchief to C, so C wins 2 supports and he is choosen to be the Hawk.

Input

There are several test cases. First is a integer T(T <= 50), means the number of test cases.

Each test case start with two integer n, m in a line (2 <= n <= 5000, 0 <m <= 30000). n means there are n children(numbered from 0 to n - 1). Each of the following m lines contains two integers A and B(A != B) denoting that the child numbered A give a handkerchief
to B.

Output

For each test case, the output should first contain one line with "Case x:", here x means the case number start from 1. Followed by one number which is the totalsupports the winner(s) get.

Then follow a line contain all the Hawks‘ number. The numbers must be listed in increasing order and separated by single spaces.

Sample Input

2
4 3
3 2
2 0
2 1

3 3
1 0
2 1
0 2

Sample Output

Case 1: 2
0 1
Case 2: 2
0 1 2

Author

Dragon

Source

2010 ACM-ICPC Multi-University
Training Contest（19）——Host by HDU

解题思路：首先是强连通分量缩点，然后重新构图，对于同一个强连通分量的点的支持度一是强连通分量内的其他点，还有其他的强连通分量对他所在强连通分量的支持，为了找到这个我们之前构图时反向建图，然后bfs

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stack>
#include<queue>
using namespace std;
#define Max 5005
int head[Max],head2[Max],j,k,bcnt,dindex,dfn[Max],low[Max],belong[Max],sup[Max];
bool visit[Max],vis[Max];;
stack<int> s;
struct
{
    int s;
    int e;
    int next;
}edge[Max*6],edge2[Max*6];
void add(int s,int e)
{
    edge[k].s=s;
    edge[k].e=e;
    edge[k].next=head[s];
    head[s]=k++;
}
void add2(int s,int e)
{
    edge2[j].s=s;
    edge2[j].e=e;
    edge2[j].next=head2[s];
    head2[s]=j++;
}
void tarjan(int i)
{
    int ed;
    dfn[i]=low[i]=++dindex;
    visit[i]=true;
    s.push(i);
    for(int t=head[i];t!=-1;t=edge[t].next)
    {
        ed=edge[t].e;
        if(!dfn[ed])
        {
            tarjan(ed);
            if(low[i]>low[ed])
                low[i]=low[ed];
        }
        else if(visit[ed]&&low[i]>dfn[ed])
            low[i]=dfn[ed];
    }
    if(dfn[i]==low[i])
    {
        bcnt++;
        do
        {
            ed=s.top();
            s.pop();
            visit[ed]=false;
            belong[ed]=bcnt;
        }while(i!=ed);
    }
}
void solve(int n)
{
    int i;
    bcnt=dindex=0;
    memset(visit,false,sizeof(visit));
    memset(low,0,sizeof(low));
    memset(belong,0,sizeof(belong));
    memset(dfn,0,sizeof(dfn));
    while(!s.empty())
        s.pop();
    for(i=0;i<n;i++)
        if(!dfn[i])
            tarjan(i);
}
int bfs(int rt)
{
    int i,st,ed,sum=0;
    vis[rt]=true;
    queue<int> q;
    q.push(rt);
    while(!q.empty())
    {
        st=q.front();
        q.pop();
        for(i=head2[st];i!=-1;i=edge2[i].next)
        {
            ed=edge2[i].e;
            if(!vis[ed])
            {
                sum+=sup[ed]+1;
                q.push(ed);
                vis[ed]=true;
            }
        }
    }
    return sum;
}
int main()
{
    int i,t,m,n,a,b,ans[Max],sum[Max],flag,cnt,p[Max],ncase=1;
    scanf("%d",&t);
    while(t--)
    {
        flag=-1;
        j=k=cnt=0;
        memset(p,0,sizeof(p));
        memset(ans,0,sizeof(ans));
        memset(sum,0,sizeof(sum));
        memset(head,-1,sizeof(head));
        memset(head2,-1,sizeof(head2));
        memset(sup,0,sizeof(sup));
        scanf("%d%d",&n,&m);
        while(m--)
        {
            scanf("%d%d",&a,&b);
            add(a,b);
        }
        solve(n);
        for(i=0;i<k;i++)
        {
            int st=edge[i].s;
            int ed=edge[i].e;
            if(belong[st]!=belong[ed])
                add2(belong[ed],belong[st]);
        }
        for(i=0;i<n;i++)
            sup[belong[i]]++;
        for(i=1;i<=bcnt;i++)
            sup[i]--;
        for(i=1;i<=bcnt;i++)
        {
            memset(vis,false,sizeof(vis));
            sum[i]=bfs(i)+sup[i];
        }
        for(i=0;i<n;i++)
        {
            ans[i]=sum[belong[i]];
            flag=max(flag,ans[i]);
        }
        for(i=0;i<n;i++)
        {
            if(ans[i]==flag)
                p[cnt++]=i;
            //cout<<ans[i]<<endl;
        }
        printf("Case %d: %d\n",ncase++,flag);
        for(i=0;i<cnt;i++)
            printf("%d%c",p[i],i==cnt-1?'\n':' ');
    }
    return 0;
}