POJ2186 Popular Cows (Tarjan)

题目链接：http://poj.org/problem?id=2186

题解：强连通分量缩点，Your task is to compute the number of cows that are considered popular by every other cow.
只需计算出度为0的点数，如果出度为0的点数大于一，输出0（不符合题意）,否则输出出度为0的强连通分量的点数。

#include <stdio.h>
#include <string.h>
#define MAXN 10001

struct node
{
int to,next;
}edge[50001];

int belong[MAXN],num[MAXN],outdegree[MAXN];
int head[MAXN],instack[MAXN],low[MAXN],dfn[MAXN];
int stack[MAXN],tot,Dindex,top,Bcnt,n;

void Init()
{//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(outdegree,0,sizeof(outdegree));
}

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);
}
}

int main()
{
int m,i,j,x;
while(scanf("%d %d",&n,&m)!=EOF)
{
Init();
while(m--)
{
i=Scan();
j=Scan();
addEdge(i,j);
}
for(i=1;i<=n;++i)
{
if(!dfn[i])
Tarjan(i);
}
for(i=1;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
{
outdegree[belong[i]]++;
//indegree[belong[x]]++;
}
}
}
x=0;
for(i=1;i<=Bcnt;++i)
{
if(outdegree[i]==0)
{
x++;
j=i;
}
}
if(x==1)
printf("%d\n",num[j]);
else
printf("0\n");
}
return 0;
}