强连通 反向建图 hdu3639

Posted 2020-09-30 Billyshuai

Hawk-and-Chicken

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3321    Accepted Submission(s): 1041

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 total supports 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

 

反向建图的原因是方便统计，倘若是正向的话 ，那么如果有2->3,3->1,2->1  2这个节点在计算的时候很难搞

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=5008;
const int M=30008;
int head[N],tot,scnt,cnt,cont,sz[N];
int dfn[N],low[N],bl[N],q[N],l;
bool instack[N],ru[N];
vector<int>G;
vector<int>C[N];
struct node{
   int to,next;
}e[M];
void add(int u,int v){
   e[tot].to=v;
   e[tot].next=head[u];
   head[u]=tot++;
}
void init(){
   tot=scnt=cnt=l=0;
   memset(head,-1,sizeof(head));
   memset(dfn,0,sizeof(dfn));
   memset(instack,0,sizeof(instack));
   memset(ru,0,sizeof(ru));
   G.clear();
   for(int i=0;i<N;++i) C[i].clear();
}
void Tajan(int u){
   dfn[u]=low[u]=++cnt;
   q[l++]=u;
   instack[u]=1;
   for(int i=head[u];i+1;i=e[i].next){
    int v=e[i].to;
    if(!dfn[v]) {
        Tajan(v);
        low[u]=min(low[u],low[v]);
    }
    else if(instack[v]&&dfn[v]<low[u])
        low[u]=dfn[v];
   }
   if(dfn[u]==low[u]) {
    ++scnt;
    int t;
    sz[scnt]=0;
    do{
        t=q[--l];
        bl[t]=scnt;
        ++sz[scnt];
        C[scnt].push_back(t);
        instack[t]=0;
    }while(t!=u);
   }
}
bool used[N];
void dfs(int u,int tz){
   for(int i=head[u];i+1;i=e[i].next)
   if(!used[e[i].to]){
    used[e[i].to]=1;
    sz[tz]+=sz[e[i].to];
    dfs(e[i].to,tz);
   }
}
struct point{
   int u,v;
}ee[M];
int main(){
    int n,m,u,v,T;
    scanf("%d",&T);
    for(int tas=1;tas<=T;++tas){
        init();
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;++i){
            scanf("%d%d",&u,&v);
            add(v,u);
        }
        for(int i=0;i<n;++i) if(!dfn[i]) Tajan(i);
        int pp=0;
        for(int i=0;i<n;++i)
        for(int j=head[i];j+1;j=e[j].next){
            int v=e[j].to;
            if(bl[i]==bl[v]) continue;
            else {ee[pp].u=bl[i];ee[pp++].v=bl[v];ru[bl[v]]=1;}
        }
        memset(head,-1,sizeof(head));
        tot=0;
        for(int i=0;i<pp;++i) add(ee[i].u,ee[i].v);
        int maxx=-1;
        for(int i=1;i<=scnt;++i)
        if(!ru[i]){
            memset(used,0,sizeof(used));
            dfs(i,i);
            if(maxx<sz[i]) {maxx=sz[i];G.clear(); for(int j=0;j<(int)C[i].size();++j) G.push_back(C[i][j]);}
        else if(maxx==sz[i]) for(int j=0;j<(int)C[i].size();++j) G.push_back(C[i][j]);
        }
        sort(G.begin(),G.end());
        printf("Case %d: %d\n",tas,maxx-1);
        for(int i=0;i<(int)G.size()-1;++i) printf("%d ",G[i]);
        printf("%d\n",G[(int)G.size()-1]);
    }
}