Popular Cows (POJ No.2186)

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



题解：
考虑以牛为顶点的有向图，对每一个需对(A,B)连一条从A到B的边。咱们不妨假设两头牛A,B都被其余牛认为是红牛。那么就知道A,B必定同属一个强连通份量，即存在一个包含A，B两个顶点的圈。反之，若是一个牛被其余牛认为是红牛，那么他所属的强连通份量中的牛必定所有是红牛。因此咱们只须要找出拓扑序最大的强连通份量的个数就能够了。

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\n",num);
77 } 

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