Popular Cows

Description

Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.

Input

* Line 1: Two space-separated integers, N and M * Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.

Output

* Line 1: A single integer that is the number of cows who are considered popular by every other cow.

Sample Input

3 3
1 2
2 1
2 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity.

Source

先縮點，如果只有一個出度為0的點，才存在符合題意的牛，數量為該強聯通分量裡的點

特判只有一個強聯通分量的情況，直接輸出n

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int MAXN = 10005;
const int MAXM = 50005;
const int INF = 0x3f3f3f3f;
/***************************************/
struct Edge
{
    int to,Next;
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int Index,top;
int scc;
bool Instack[MAXN];
int num[MAXN];
void addedge(int u,int v)
{
    edge[tot].to = v;
    edge[tot].Next = head[u];
    head[u] = tot++;
}
void Tarjan(int u)
{
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = head[u]; i != -1; i = edge[i].Next) {
        v = edge[i].to;
        if(!DFN[v]) {
            Tarjan(v);
            if(Low[u] > Low[v]) Low[u] = Low[v];
        }
        else if(Instack[v] && Low[u] > DFN[v])
            Low[u] = DFN[v];
    }
    if(Low[u] == DFN[u]) {
        scc++;
        do {
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc;
            num[scc]++;
        }
        while(v != u);
    }
}
void solve(int N)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(num,0,sizeof(num));
    Index = scc = top = 0;
    for(int i = 1; i <= N; i++) {
        if(!DFN[i]) {
            Tarjan(i);
        }
    }
}
int in[MAXN],out[MAXN];
void init()
{
    tot = 0;
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    memset(head,-1,sizeof(head));
}
/***************************************/
int main(void)
{
    int n,m,u,v;
    scanf("%d %d",&n,&m);
    init();
    for(int i = 1; i <= m; i++) {
        scanf("%d %d",&u,&v);
        addedge(u,v);
    }
    solve(n);
    if(scc == 1) {
        printf("%d\n",n);
        return 0;
    }
    for(int i = 1; i <= n; i++) {
        for(int j = head[i]; j != -1; j = edge[j].Next) {
            u = Belong[i],v = Belong[edge[j].to];
            if(u != v) out[u]++,in[v]++;
        }
    }
    int cnt = 0,ans;
    for(int i = 1; i <= scc; i++) {
        if(out[i] == 0) {
            cnt++;
            ans = num[i];
        }
    }
    if(cnt == 1) printf("%d\n",ans);
    else printf("0\n");
    return 0;
}