1 /**
2 problem: http://poj.org/problem?id=2186
3
4 当出度为0的点（可能是缩点后的点）只有一个时就存在被所有牛崇拜的牛
5 因为如果存在有两个及以上出度为0的点的话，他们不会互相崇拜所以不存在被所有牛崇拜的牛
6 牛的个数即该出度为0点（由环缩点而来）有多少牛
7 **/
8 #include<stdio.h>
9 #include<stack>
10 #include<algorithm>
11 using namespace std;
12
13 class Graphics{
14 const static int MAXN = 10005;
15 const static int MAXM = 50005;
16 private:
17 struct Edge{
18 int to, next;
19 }edge[MAXM];
20 struct Point{
21 int dfn, low, color;
22 }point[MAXN];
23 int sign, dfnNum, colorNum, sumOfPoint, first[MAXN], count[MAXN];
24 bool vis[MAXN];
25 stack<int> stk;
26 void tarjan(int u){
27 point[u].dfn = ++dfnNum;
28 point[u].low = dfnNum;
29 vis[u] = true;
30 stk.push(u);
31 for(int i = first[u]; i != -1; i = edge[i].next){
32 int to = edge[i].to;
33 if(!point[to].dfn){
34 tarjan(to);
35 point[u].low = min(point[to].low, point[u].low);
36 }else if(vis[to]){
37 point[u].low = min(point[to].dfn, point[u].low);
38 }
39 }
40 if(point[u].dfn == point[u].low){
41 vis[u] = false;
42 point[u].color = ++colorNum;
43 count[colorNum] ++;
44 while(stk.top() != u){
45 vis[stk.top()] = false;
46 point[stk.top()].color = colorNum;
47 count[colorNum] ++;
48 stk.pop();
49 }
50 stk.pop();
51 }
52 }
53 public:
54 void clear(int n){
55 sign = dfnNum = colorNum = 0;
56 for(int i = 1; i <= n; i ++){
57 first[i] = -1;
58 vis[i] = 0;
59 count[i] = 0;
60 }
61 sumOfPoint = n;
62 while(!stk.empty()) stk.pop();
63 }
64 void addEdgeOneWay(int u, int v){
65 edge[sign].to = v;
66 edge[sign].next = first[u];
67 first[u] = sign ++;
68 }
69 void addEdgeTwoWay(int u, int v){
70 addEdgeOneWay(u, v);
71 addEdgeOneWay(v, u);
72 }
73 void tarjanAllPoint(){
74 for(int i = 1; i <= sumOfPoint; i ++){
75 if(!point[i].dfn)
76 tarjan(i);
77 }
78 }
79 int getAns(){
80 int ans = 0, ans2 = 0;
81 int *outdegree = new int[sumOfPoint+1];
82 for(int i = 1; i <= sumOfPoint; i ++){
83 outdegree[i] = 0;
84 }
85 tarjanAllPoint();
86 for(int i = 1; i <= sumOfPoint; i ++){
87 for(int j = first[i]; j != -1; j = edge[j].next){
88 int to = edge[j].to;
89 if(point[to].color != point[i].color){
90 outdegree[point[i].color] ++;
91 }
92 }
93 }
94 for(int i = 1; i <= colorNum; i ++){
95 if(!outdegree[i]){
96 ans ++;
97 ans2 = count[i];
98 }
99 }
100 delete []outdegree;
101 if(ans == 1){
102 return ans2;
103 }else{
104 return 0;
105 }
106 }
107 }graph;
108
109 int main(){
110 int n, m;
111 scanf("%d%d", &n, &m);
112 graph.clear(n);
113 while(m --){
114 int a, b;
115 scanf("%d%d", &a, &b);
116 graph.addEdgeOneWay(a, b);
117 }
118 printf("%d\n", graph.getAns());
119 return 0;
120 }
