Popular cow - Los Valley 2341

Gugu Gu

 

(Likely to stay when the winter training camp to the bar, the result was I mutter to now ...

Tarjan simple application, which is very simple to see.

FIG star cows then only one of the components is a strong link 0.

If there are two or more of the components of a strong link 0, then there is no star cows, because these can not pass the degree of "love" strongly connected components between 0

So, the final answer is the number of cows Unicom strong component in!

#include<cstdio>
#include<algorithm>
#include<queue> 
#include<stack>
#define ll long long
using namespace std;

inline ll read()
{
    ll sum = 0, p = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')
            p = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        (sum *= 10) += ch - '0';
        ch = getchar();
    }
    return sum * p;
}

const int maxn = 10005,maxm = 50005;
struct edge
{
    int nxt,to;
}e[maxm];
int n,m,cnt,tot,ncnt;
int dfn[maxn],low[maxn],mrk[maxn],id[maxn],du[maxn],head[maxn];
bool vis[maxn];
stack<int> q;

void add(int a,int b)
{
    e[++cnt].to = b;
    e[cnt].nxt = head[a];
    head[a] = cnt;
}

void tarjan(int x)
{
    dfn[x] = low[x] = ++tot;
    q.push(x);
    vis[x] = true;
    for(int i = head[x];i;i = e[i].nxt)
    {
        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],dfn[v]);
    }
    int k;
    if(dfn[x] == low[x])
    {
        ncnt++;
        do
        {
            k = q.top();
            q.pop();
            vis[k] = false;
            id[k] = ncnt;
            mrk[ncnt]++;
        }while(x != k); 
    }
}

int main()
{
    n = read(),m = read();
    for(int i = 1;i <= m;i++)
    {
        int a = read(),b = read();
        add(a,b);
    }
    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;j = e[j].nxt)
        {
            int v = e[j].to;
            if(id[i] != id[v])
                du[id[i]]++;
        }
    int ans = 0;
    for(int i = 1;i <= ncnt;i++)
    if(!du[i])
    {
        if(ans)
        {
            printf("0\n");
            return 0;
        }
        ans = i;
    }
    printf("%d\n",mrk[ans]);
    return 0;
}