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Hawk-and-Chicken

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1979 Accepted Submission(s): 570


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 解題思路：首先是強連通分量縮點，然後又一次構圖，對於同一個強連通分量的點的支持度一是強連通分量內的其它點。還有其它的強連通分量對他所在強連通分量的支持，為了找到這個我們之前構圖時反向建圖，然後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;
}

HDU 3639 Hawk-and-Chicken