题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=3639

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

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题目意思：

给一幅图，求点i,使得能够到达i的点数最多。

解题思路：

tarjan缩点，反向建图，然后dfs统计入度为0的联通分量，求出能够到达的节点数a，比较记录a最大的那些连通分量。

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 5500

int low[Maxn],dfn[Maxn],sta[Maxn],sc,bc,n,m,dep;
bool iss[Maxn];
int in[Maxn],nu[Maxn],dei[Maxn],ans,temp;
vector<vector<int> >myv;
vector<vector<int> >tree;
map<int,int>myp[Maxn];
bool vis[Maxn];

void tarjan(int cur)
{
    int ne;

    low[cur]=dfn[cur]=++dep;
    sta[++sc]=cur;
    iss[cur]=true;

    for(int i=0;i<myv[cur].size();i++)
    {
        ne=myv[cur][i];
        if(!dfn[ne])
        {
            tarjan(ne);
            if(low[ne]<low[cur])
                low[cur]=low[ne];
        }
        else if(iss[ne]&&dfn[ne]<low[cur])
            low[cur]=dfn[ne];
    }
    if(low[cur]==dfn[cur])
    {
        bc++;
        do
        {
            ne=sta[sc--];
            iss[ne]=false;
            in[ne]=bc;
            nu[bc]++;
        }while(ne!=cur);
    }
}
void solve()
{
    dep=sc=bc=0;
    memset(iss,false,sizeof(iss));
    memset(dfn,0,sizeof(dfn));
    memset(nu,0,sizeof(nu));

    for(int i=0;i<n;i++)
        if(!dfn[i])
            tarjan(i);
}
void dfs(int cur)
{
    vis[cur]=true;
    temp+=nu[cur];

    for(int j=0;j<tree[cur].size();j++)
    {
        int ne=tree[cur][j];
        if(!vis[ne])
            dfs(ne);
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   int t,kcas=0;

   scanf("%d",&t);
   while(t--)
   {
       scanf("%d%d",&n,&m);
       myv.clear();
       myv.resize(n+1);

       for(int i=1;i<=m;i++)
       {
           int a,b;
           scanf("%d%d",&a,&b);
           myv[a].push_back(b);
       }
       solve();

       tree.clear();
       tree.resize(bc+1);
       for(int i=1;i<=bc;i++)
            myp[i].clear();

       memset(dei,0,sizeof(dei));
       for(int i=0;i<n;i++)
       {
           for(int j=0;j<myv[i].size();j++)
           {
               int ne=myv[i][j];
               if(in[ne]!=in[i])
               {
                   if(myp[in[ne]][in[i]]==0)
                   {
                       //system("pause");
                       myp[in[ne]][in[i]]=1;
                       tree[in[ne]].push_back(in[i]);
                       dei[in[i]]++;
                   }
               }
           }
       }
       ans=0;
       /*printf("bc:%d\n",bc);
       for(int i=0;i<n;i++)
            printf("i:%d bei:%d dei:%d\n",i,in[i],dei[in[i]]);*/
       for(int i=1;i<=bc;i++)
       {
           if(!dei[i])
            {
                //printf("i:%d\n",i);
                //system("pause");

                for(int j=1;j<=bc;j++)
                    vis[j]=false;
                temp=0;
                dfs(i);
                nu[i]=temp;
                ans=max(ans,temp);
            }
       }
       printf("Case %d: %d\n",++kcas,ans-1);
       bool fi=true;

       for(int i=0;i<n;i++)
       {
           if(nu[in[i]]==ans)
           {
               if(!fi)
                    putchar(' ');
               else
                    fi=false;
               printf("%d",i);
           }

       }
       putchar('\n');
   }

    return 0;
}
/*
4 4
0 1
0 2
1 3
2 3
*/