【luogu P2746 [USACO5.3]校园网Network of Schools】 题解

题目链接：https://www.luogu.org/problemnew/show/P2812
注意：判断出入度是否为0的时候枚举只需到颜色的数量。
坑点：当只有一个强连通分量时，不需要再添加新边。即子任务B ans = 0。
子任务B证明：若每个点都相连通，出入度都必须为1。
保证所有点的出入度都>1就OK。
所以需要找一下出度为0和入度为0的点再取一个max即可。

#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 10;
struct edge{
    int from, to, next;
}e[maxn<<2];
int head[maxn], cnt;
bool vis[maxn];
int n, dfn[maxn], low[maxn], tim, color[maxn], num, chudu[maxn], rudu[maxn], runum, chunum;

stack<int> s;
void add(int u, int v)
{
    e[++cnt].from = u;
    e[cnt].next = head[u];
    e[cnt].to = v;
    head[u] = cnt;
}
void tarjan(int x)
{
    dfn[x] = low[x] = ++tim;
    s.push(x); vis[x] = 1;
    for(int i = head[x]; i != -1; i = e[i].next)
    {
        int v = e[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            low[x] = min(low[x], low[v]);
        }
        else if(vis[v])
        {
            low[x] = min(low[x], low[v]);
        }
    }
    if(dfn[x] == low[x])
    {
        color[x] = ++num;
        vis[x] = 0;
        while(s.top() != x)
        {
            color[s.top()] = num;
            vis[s.top()] = 0;
            s.pop();
        }
        s.pop();
    }
}
int main()
{
    int m;
    memset(head, -1, sizeof(head));
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    {
        int u;
        while(scanf("%d",&u) && u != 0)
        add(i, u);
    }
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) tarjan(i);
    for(int i = 1; i <= n; i++)
        for(int j = head[i]; j != -1; j = e[j].next)
        {
            int v = e[j].to;
            if(color[v] != color[i])
            {
                chudu[color[i]]++;
                rudu[color[v]]++;
            }
        }
    for(int i = 1; i <= num; i++)
    {
        if(!rudu[i]) runum++;
        if(!chudu[i]) chunum++;
    }
    for(int i = 2; i <= n; i++)
    {
        if(color[i] != color[i-1])
        {
            printf("%d\n%d",runum, max(runum, chunum));
            return 0;
        }
    }
    printf("%d\n%d",runum,0);   
    return 0;
}