hdu3639 Hawk-and-Chicken (Tarjan)

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

题解：将强连通分量缩成一个点,反向建图，对每个入度为0的点DFS,找出最大值即可。

#include <cstdio>        
#include <cstring> 
#include <vector>
#define MAXN 5001   
using namespace std;

struct node        
{        
	int to,next; 
}edge[30001];    

vector<int>mp[MAXN];
int belong[MAXN],num[MAXN],maxans[MAXN];        
int head[MAXN],instack[MAXN],low[MAXN],dfn[MAXN];          
int stack[MAXN],tot,Dindex,top,Bcnt,n,indegree[MAXN];    

void Init(int n)          
{//初始化   
	int i;  
	tot=0,top=0,Dindex=0,Bcnt=0;          
	memset(head,-1,sizeof(head));          
	memset(instack,0,sizeof(instack));          
	memset(dfn,0,sizeof(dfn));          
	memset(low,0,sizeof(low));         
	memset(belong,0,sizeof(belong));  
	memset(num,0,sizeof(num));
	memset(indegree,0,sizeof(indegree));
	memset(maxans,0,sizeof(maxans));
	for(i=0;i<=n;++i)
		mp[i].clear();
} 

void addEdge(int from,int to)          
{        
	edge[tot].to=to;
	edge[tot].next=head[from];          
	head[from]=tot++;          
}   

int Scan()                  
{                  
	char ch;                  
	int ret=0;                  
	while((ch=getchar())<'0'||ch>'9');                  
	while(ch>='0'&&ch<='9')                  
	{                  
		ret=ret*10+(ch-'0');                  
		ch=getchar();                  
	}                  
	return ret;                  
} 

void Tarjan(int x)        
{        
	int i,u,v;        
	dfn[x]=low[x]=++Dindex;//时间戳        
	stack[top++]=x;        
	instack[x]=1;        
	for(i=head[x];i!=-1;i=edge[i].next)        
	{        
		u=edge[i].to;        
		if(!dfn[u])        
		{        
			Tarjan(u);        
			low[x]=low[x]>low[u]?low[u]:low[x];        
		}        
		else if(instack[u]&&low[x]>dfn[u])        
			low[x]=dfn[u];        
	}        
	if(low[x]==dfn[x])        
	{   
        //Bcnt++;        
		do         
		{        
			v=stack[--top];        
			instack[v]=0;     
			belong[v]=Bcnt; 
			num[Bcnt]++;//保存该强连通分量的点数
		} while (v!=x); 
		Bcnt++;
	}        
}   

void DFS(int x)
{
	instack[x]=1;
	tot+=num[x];
	for(int i=0;i<mp[x].size();++i)
	{
		if(!instack[mp[x][i]])
			DFS(mp[x][i]);
	}
	return ;
}

int main()
{
	int m,test,i,j,x,Max,cases=1;
	scanf("%d",&test);
	while(test--)
	{
		n=Scan();
		m=Scan();
		Init(n);
		while(m--)
		{
			i=Scan();
			j=Scan();
		//	i++,j++;
			addEdge(i,j);
		}
		for(i=0;i<n;++i)
		{
			if(!dfn[i])
				Tarjan(i);
		}
		for(i=0;i<n;++i)
		{
			for(j=head[i];j!=-1;j=edge[j].next)  
			{  
				x=edge[j].to;  
				if(belong[i]!=belong[x]) //不在同一个连通分量的点
				{  
					mp[belong[x]].push_back(belong[i]);
					indegree[belong[i]]++;
				}  
			} 
		}
		Max=-1;
		for(i=0;i<Bcnt;++i)
		{
			tot=0;
			if(indegree[i]==0)
			{
				memset(instack,0,sizeof(instack));
				DFS(i);
				maxans[i]=tot;//保存每一个入度为0的最大值
				Max=Max<tot?tot:Max;
			}
		}
		printf("Case %d: %d\n",cases++,Max-1);
		x=0;
		for(i=0;i<n;++i)
		{
			if(maxans[belong[i]]==Max)
			{
				if(!x)
				{
					printf("%d",i);
					x=1;
				}
				else
					printf(" %d",i);
			}
		}
		printf("\n");
	}
	return 0;
}