题目大意：有光盘可以传着看，问最少从哪几个人分发，能全部传一遍。
题解：缩点后求入度为0的点的个数
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 22000
using namespace std; int n,sumedge,sumclr,top,tim,ans;
int Stack[maxn],instack[maxn],low[maxn],dfn[maxn],bel[maxn],rd[maxn],head[maxn]; struct Edge{
int x,y,nxt;
Edge(int x=,int y=,int nxt=):
x(x),y(y),nxt(nxt){}
}edge[maxn<<]; void add(int x,int y){
edge[++sumedge]=Edge(x,y,head[x]);
head[x]=sumedge;
} void Tarjian(int x){
Stack[++top]=x;instack[x]=true;
low[x]=dfn[x]=++tim;
for(int i=head[x];i;i=edge[i].nxt){
int v=edge[i].y;
if(instack[v])low[x]=min(low[x],dfn[v]);
else if(!dfn[v]){
Tarjian(v);
low[x]=min(low[x],low[v]);
}
}
if(low[x]==dfn[x]){
sumclr++;
while(Stack[top+]!=x){
bel[Stack[top]]=sumclr;
instack[Stack[top]]=false;
top--;
}
}
} int main(){
scanf("%d",&n);
for(int i=;i<=n;i++){
int x;
while(){
scanf("%d",&x);
if(!x)break;
add(i,x);
}
}
for(int i=;i<=n;i++)if(!dfn[i])Tarjian(i);
for(int x=;x<=n;x++){
for(int i=head[x];i;i=edge[i].nxt){
int v=edge[i].y;
if(bel[x]!=bel[v])rd[bel[v]]++;
}
}
for(int i=;i<=sumclr;i++)if(!rd[i])ans++;
printf("%d\n",ans);
return ;
}
AC