Popular Cows
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 37083 | Accepted: 15104 |
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
Hint
Cow 3 is the only cow of high popularity.
Source
题:A欢迎B B欢迎C 则A也欢迎C 问 有多少只牛被出自己之外的所有的牛欢迎
强联通图进行缩点 当然联通块只能有一个，否则输出0
缩点之后 出度为0的点的个数就是答案
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cctype> 5 #include<cmath> 6 #include<cstring> 7 #include<map> 8 #include<queue> 9 #include<stack> 10 #include<set> 11 #include<vector> 12 #include<algorithm> 13 #include<string.h> 14 typedef long long ll; 15 typedef unsigned long long LL; 16 using namespace std; 17 const int INF=0x3f3f3f3f; 18 const double eps=0.0000000001; 19 const int N=30000+10; 20 const ll mod=1e9+7; 21 int dfn[N]; 22 int low[N]; 23 int vis[N]; 24 int head[N]; 25 int cnt; 26 stack<int>st; 27 int belong[N]; 28 struct node{ 29 int to,next; 30 }edge[N<<1]; 31 int num[N]; 32 int t,tot; 33 int in[N]; 34 int out[N]; 35 void init(){ 36 memset(head,-1,sizeof(head)); 37 memset(low,0,sizeof(low)); 38 memset(belong,-1,sizeof(belong)); 39 memset(vis,0,sizeof(vis)); 40 memset(in,0,sizeof(in)); 41 memset(out,0,sizeof(out)); 42 memset(num,0,sizeof(num)); 43 tot=t=0; 44 cnt=0; 45 } 46 void add(int u,int v){ 47 edge[tot].to=v; 48 edge[tot].next=head[u]; 49 head[u]=tot++; 50 } 51 void tarjan(int u){ 52 low[u]=dfn[u]=++t; 53 vis[u]=1; 54 st.push(u); 55 for(int i=head[u];i!=-1;i=edge[i].next){ 56 int v=edge[i].to; 57 if(dfn[v]==0){ 58 tarjan(v); 59 low[u]=min(low[u],low[v]); 60 } 61 else if(vis[v]){ 62 low[u]=min(low[u],dfn[v]); 63 } 64 } 65 if(low[u]==dfn[u]){ 66 int vv; 67 cnt++; 68 do{ 69 vv=st.top(); 70 st.pop(); 71 belong[vv]=cnt; 72 num[cnt]++; 73 vis[vv]=0; 74 }while(vv!=u); 75 } 76 } 77 int main(){ 78 int n,m; 79 while(scanf("%d%d",&n,&m)!=EOF){ 80 init(); 81 for(int i=1;i<=m;i++){ 82 int u,v; 83 scanf("%d%d",&u,&v); 84 add(u,v); 85 } 86 for(int i=1;i<=n;i++){ 87 if(dfn[i]==0)tarjan(i); 88 } 89 for(int i=1;i<=n;i++){ 90 for(int j=head[i];j!=-1;j=edge[j].next){ 91 int v=edge[j].to; 92 if(belong[i]!=belong[v]){ 93 in[belong[v]]++; 94 out[belong[i]]++; 95 } 96 } 97 } 98 int flag=0; 99 int ans=0; 100 //cout<<cnt<<endl; 101 for(int i=1;i<=cnt;i++){ 102 if(out[i]==0){ 103 flag++;ans=ans+num[i]; 104 } 105 } 106 if(flag==1){ 107 cout<<ans<<endl; 108 } 109 else{ 110 cout<<0<<endl; 111 } 112 } 113 }