题目链接：http://codeforces.com/contest/505/problem/D
题解：先用tarjan缩点然后再用并查集注意下面这种情况
‘
这种情况只需要构成一个大环就行了，也就是说不需要7条边只要6条就够了。
#include
#include
#include
#include
using namespace std;
const int M = 2e6 + 10;
struct Edge {
int v , next;
}edge[M];
int head[M] , e;
int Low[M] , DFN[M] , Stack[M] , Belong[M];
int scc;
int Index , top;
bool Instack[M];
int num[M];
void init() {
memset(head , -1 , sizeof(head));
e = 0;
}
void add(int u , int v) {
edge[e].v = v , edge[e].next = head[u] , head[u] = e++;
}
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) {
int v = edge[i].v;
if(!DFN[v]) {
Tarjan(v);
Low[u] = min(Low[u] , Low[v]);
}
else if(Instack[v]) Low[u] = min(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);
}
}
int f[M] , cnt[M] , flag[M];
int find(int x) {
if(x == f[x]) return x;
int tmp = find(f[x]);
return f[x] = tmp;
}
int main() {
int n , m;
scanf("%d%d" , &n , &m);
init();
for(int i = 0 ; i < m ; i++) {
int u , v;
scanf("%d%d" , &u , &v);
add(u , v);
}
memset(DFN , 0 , sizeof(DFN));
memset(Instack , false , sizeof(Instack));
memset(num , 0 , sizeof(num));
memset(cnt , 0 , sizeof(cnt));
memset(flag , 0 , sizeof(flag));
scc = 0 , top = 0 , Index = 0;
for(int i = 1 ; i <= n ; i++)
if(!DFN[i]) Tarjan(i);
for(int i = 1 ; i <= scc ; i++) {f[i] = i; if(num[i] > 1) flag[i] = 1;}
for(int i = 1 ; i <= n ; i++) {
for(int j = head[i] ; j != -1 ; j = edge[j].next) {
int v = edge[j].v;
if(Belong[i] != Belong[v]) {
int a = Belong[i] , b = Belong[v];
if(a != b) {
int t1 = find(a) , t2 = find(b);
if(t1 != t2) {
f[t2] = t1;
if(num[t2] > 1 || num[t1] > 1) flag[t1] = 1;
flag[t1] += flag[t2];
}
}
}
}
}
int ans = 0;
for(int i = 1 ; i <= scc ; i++) {
ans += num[i];
f[i] = find(i);
if(f[i] == i)
if(!flag[i]) ans--;
}
printf("%d\n" , ans);
return 0;
}