洛谷P2835 刻录光盘

传送门

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=0,int y=0,int nxt=0):
        x(x),y(y),nxt(nxt){}
}edge[maxn<<1];

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+1]!=x){
            bel[Stack[top]]=sumclr;
            instack[Stack[top]]=false;
            top--;
        }
    }
}

int main(){
    scanf("%d",&n);
for(int i=1;i<=n;i++){
int x;
while(1){
            scanf("%d",&x);
if(!x)break;
            add(i,x);
        }
    }
for(int i=1;i<=n;i++)if(!dfn[i])Tarjian(i);
for(int x=1;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=1;i<=sumclr;i++)if(!rd[i])ans++;
    printf("%d\n",ans);
return 0;
}