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.
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.
* 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
题解：
考虑以牛为顶点的有向图，对每个需对(A,B)连一条从A到B的边。我们不妨假设两头牛A,B都被其他牛认为是红牛。那么就知道A,B一定同属一个强连通分量，即存在一个包含A，B两个顶点的圈。反之，如果一个牛被其他牛认为是红牛，那么他所属的强连通分量中的牛一定全部是红牛。所以我们只需要找出拓扑序最大的强连通分量的个数就可以了。
AC代码：
1 #include<iostream> 2 #include<cctype> 3 using namespace std; 4 const int MAXN=500000+10; 5 //------------------------- 6 void read(int &x){ 7 x=0;char ch=getchar();int f=1; 8 for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 9 for(;isdigit(ch);ch=getchar())x=x*10+ch-'0'; 10 x*=f; 11 } 12 //------------------------- 13 int n,m,tmp; 14 int topo[MAXN],cmp[MAXN]; 15 bool vis[MAXN]; 16 int first[MAXN],next[MAXN],v[MAXN],e; 17 void AddEdge(int a,int b){ 18 v[++e]=b; 19 next[e]=first[a]; 20 first[a]=e; 21 } 22 23 int rfirst[MAXN],rnext[MAXN],rv[MAXN],re; 24 void rAddEdge(int a,int b){ 25 rv[++re]=b; 26 rnext[re]=rfirst[a]; 27 rfirst[a]=re; 28 } 29 //------------------------- 30 void dfs(int x){ 31 vis[x]=1; 32 for(int i=first[x];i;i=next[i]) 33 if(!vis[v[i]])dfs(v[i]); 34 topo[++tmp]=x; 35 } 36 37 void rdfs(int x,int k){ 38 vis[x]=1; 39 cmp[x]=k; 40 for(int i=rfirst[x];i;i=rnext[i]) 41 if(!vis[rv[i]])rdfs(rv[i],k); 42 } 43 //--------------------------- 44 int k=1; 45 int scc(){ 46 memset(vis,0,sizeof(vis)); 47 memset(topo,0,sizeof(topo)); 48 for(int i=1;i<=n;i++){ 49 if(!vis[i])dfs(i); 50 } 51 memset(vis,0,sizeof(vis)); 52 for(int i=n;i>=1;i--)if(!vis[topo[i]])rdfs(topo[i],k++); 53 return k-1; 54 } 55 //--------------------------- 56 int main(){ 57 read(n);read(m); 58 for(int i=1;i<=m;i++){ 59 int x,y; 60 read(x);read(y); 61 AddEdge(x,y); 62 rAddEdge(y,x); 63 } 64 int nn=scc(); 65 66 int u=0,num=0; 67 for(int i=1;i<=n;i++) 68 if(cmp[i]==nn){u=i;num++;} 69 memset(vis,0,sizeof(vis)); 70 rdfs(u,0); 71 for(int i=1;i<=n;i++) 72 if(!vis[i]){ 73 num=0; 74 break; 75 } 76 printf("%d ",num); 77 }