HDU 3639 Hawk-and-Chicken, hawkandeagle

HDU 3639 Hawk-and-Chicken, hawkandeagle
Hawk-and-ChickenTime Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 1979 Accepted Submission (s): 570


Problem DescriptionKids 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 (! = B) A can win only one support from B in any case the number of the supports transmitted from B to A are supported. 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.
InputThere 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 (! = B) denoting that the child numbered A give a handkerchief to B.
OutputFor each test case, the output shoshould 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

24 33 22 02 13 31 02 10 2

 
Sample Output

Case 1: 20 1Case 2: 20 1 2

 
AuthorDragon
ACM-ICPC Multi-University Training Contest (19) -- Host by HDU Solution: first, the point of the strongly connected component is reduced and then re-formed. The support for the points of the same strongly connected component is the other points in the strongly connected component, there are other strongly connected components that support their strong connected components. In order to find this, we reverse-map the component in the previous diagram, and then bfs

#include<iostream>#include<cstdio>#include<cstring>#include<stack>#include<queue>using namespace std;#define Max 5005int 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;}