POJ 2186 Popular Cows

　　　　　　　　　　　　　　　　　　　　Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 32017		Accepted: 13039

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

Sample Output

Hint

Cow 3 is the only cow of high popularity.

Source

USACO 2003 Fall

题目大意：有很多奶牛，很多仰慕关系，仰慕关系存在传递性，求所有的牛都仰慕的对象有多少。。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<stack>
 5 #define maxn 10005
 6 #define maxm 50010
 7 using namespace std;
 8 int n,m;
 9 struct edge{
10     int u,v,next;
11 }e[maxm],ee[maxm];
12 int head[maxn],js,headd[maxn],jss,bianshu;
13 bool exist[maxn];
14 int visx,cur;// cur--缩出的点的数量 
15 int dfn[maxn],low[maxn],belong[maxn],chudu[maxn],rudu[maxn];
16 stack<int>st;
17 void add_edge(int u,int v){
18     e[++js].u=u;e[js].v=v;
19     e[js].next=head[u];head[u]=js;
20 }
21 void add_edgee(int u,int v){
22     ee[++jss].u=u;ee[jss].v=v;
23     ee[jss].next=head[u];head[u]=jss;
24 }
25 void tarjan(int u){
26     dfn[u]=low[u]=++visx;
27     st.push(u);exist[u]=true;
28     for(int i=head[u];i;i=e[i].next){
29         int v=e[i].v;
30         if(dfn[v]==0) {
31             tarjan(v);
32             low[u]=min(low[v],low[u]);
33         }
34         else if(exist[v]&&low[u]>dfn[v]) low[u]=dfn[v];
35     }
36     int j;
37     if(low[u]==dfn[u]){
38         ++cur;
39         do{
40             j=st.top();st.pop();
41             exist[j]=false;belong[j]=cur;
42             rudu[cur]++;
43         }while(j!=u);
44     }
45 }
46 int main()
47 {
48     scanf("%d%d",&n,&m);
49     int u,v;
50     for(int i=1;i<=m;i++){
51         scanf("%d%d",&u,&v);
52         add_edge(u,v);
53     }
54     for(int i=1;i<=n;i++)
55       if(dfn[i]==0)
56         tarjan(i);      
57     for(int i=1;i<=js;i++){
58         int u=e[i].u,v=e[i].v;
59         if(belong[u]!=belong[v]){
60             add_edgee(belong[u],belong[v]);
61             chudu[belong[u]]++;
62         }
63     }
64     int chu=0,ru=0;
65     for(int i=1;i<=cur;i++){
66         if(chudu[i]==0) chu++,bianshu=i;
67     }
68     if(chu==1) printf("%d
",rudu[bianshu]);
69     else printf("%d
",0);
70     return 0;
71 }

思路：首先强连通分量中的奶牛都互相仰慕，所以进行缩点，然后求出出度为零的点的个数，若有一个，则就是答案（这个强连通分量的点的个数），若不止一个，则无答案。。。//我用rudu数组记录的每个强连通分量里点的个数。。。