POJ 2186
Popular Cows
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 21823 | Accepted: 8900 |
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
ps:简单的强连通缩点。理解就好
1 //强连通
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<vector>
6 using namespace std;
7 const int MAX = 50000+10;
8 int vis[MAX]; int sccno[MAX];
9 vector<int> G[MAX],vs,RG[MAX];
10 void add_edge(int from,int to)
11 {
12 G[from].push_back(to);
13 RG[to].push_back(from);
14 }
15 void dfs(int u)
16 {
17 if(vis[u]) return ;
18 vis[u]=1;
19 for(int i=0;i<G[u].size();i++)
20 {
21 if(!vis[G[u][i]]) dfs(G[u][i]);
22 }
23 vs.push_back(u);
24 }
25 void rdfs(int u,int k)
26 {
27 if(sccno[u]) return ;
28 sccno[u]=k;
29 for(int i=0;i<RG[u].size();i++)
30 {
31 if(!sccno[RG[u][i]]) rdfs(RG[u][i],k);
32 }
33 }
34 int main()
35 {
36 int n,m; int a,b,cnt;
37 while(scanf("%d %d",&n,&m)==2)
38 {
39 for(int i=0;i<m;i++)
40 {
41 scanf("%d %d",&a,&b);
42 add_edge(a,b);
43 }
44 memset(vis,0,sizeof(vis));
45 memset(sccno,0,sizeof(sccno));
46 for(int i=1;i<=n;i++) dfs(i);
47 cnt=0;
48 for(int i=n-1;i>=0;i--)
49 {
50 if(!sccno[vs[i]])
51 {
52 cnt++;
53 rdfs(vs[i],cnt);
54 }
55 }
56 int u,ans=0;
57 for(int i=1;i<=n;i++)
58 {
59 if(sccno[i]==cnt)
60 {
61 u=i;
62 ans++;
63 }
64 }
65
66 memset(sccno,0,sizeof(sccno));
67 rdfs(u,1); int flag=1;
68 for(int i=1;i<=n;i++)
69 {
70 if(!sccno[i]) flag=0;
71 }
72 if(flag) printf("%d\n",ans);
73 else printf("0\n");
74 }
75 return 0;
76 }
