poj2186 Popular Cows 题解——S.B.S.

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Popular Cows

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 29642 Accepted: 11996

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 31 22 12 3

Sample Output

Hint

Cow 3 is the only cow of high popularity.

Source

USACO 2003 Fall

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图的强联通分量。

题目大意：

有n只牛，牛A认为牛B很牛，牛B认为牛C很牛。
给你M个关系（谁认为谁牛），求大家都认为它很牛的牛有几只。
p.s.如果牛A认为牛B很牛，牛B认为牛C很牛。那么我们就认为牛A认为牛C很牛。
  1 /*  2     Problem:   poj 2186  3     OJ:         POJ  4     User:     S.B.S.  5     Time:     297 ms  6     Memory:     1468 kb  7     Length:     2420 b  8 */  9 #include<iostream> 10 #include<cstdio> 11 #include<cstring> 12 #include<cmath> 13 #include<algorithm> 14 #include<queue> 15 #include<cstdlib> 16 #include<iomanip> 17 #include<cassert> 18 #include<climits> 19 #include<functional> 20 #include<bitset> 21 #include<vector> 22 #include<list> 23 #include<map> 24 #define maxn 10001 25 #define F(i,j,k) for(int i=j;i<=k;i++) 26 #define M(a,b) memset(a,b,sizeof(a)) 27 #define FF(i,j,k) for(int i=j;i>=k;i--) 28 #define inf 0x3f3f3f3f 29 #define maxm 50001 30 #define mod 998244353 31 //#define LOCAL 32 using namespace std; 33 int read(){ 34     int x=0,f=1;char ch=getchar(); 35     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 36     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 37     return x*f; 38 } 39 int n,m; 40 struct EDGE   41 {   42     int from; 43     int to; 44     int next;   45 }edge[maxm];   46 int head[maxn]; 47 int dfn[maxn],low[maxn],stack1[maxn]; 48 int num[maxn],du[maxn],vis[maxn];   49 int tim,tp,dcnt,sum; 50 int cnt,time=1,top,cut,tot; 51 inline void addedge(int u,int v)   52 { 53     edge[tot].from=u;   54     edge[tot].to=v;   55     edge[tot].next=head[u];   56     head[u]=tot;   57     tot++;   58 } 59 inline void dfs(int u,int fa)   60 {   61     dfn[u]=time;   62     low[u]=time;   63     time++;   64     vis[u]=1;   65     stack1[top]=u;   66     top++;   67     for(int i=head[u]; i!=-1; i=edge[i].next)   68     {   69         int v=edge[i].to;   70         if(!vis[v]){   71             dfs(v,u);   72             low[u]=min(low[u],low[v]);   73         }   74         else if(vis[v]){   75             low[u]=min(low[u],dfn[v]);   76         }   77     }   78     if(low[u]==dfn[u]){   79         cut++;   80         while(top>0&&stack1[top]!=u)   81         {   82             top--;   83             vis[stack1[top]]=2;   84             num[stack1[top]]=cut;   85         }   86     }   87 }   88 int main() 89 { 90     std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y; 91     #ifdef LOCAL 92     freopen("data.in","r",stdin); 93     freopen("data.out","w",stdout); 94     #endif 95     cin>>n>>m; 96     M(head,-1); 97     F(i,0,m-1){ 98         int a,b; 99         cin>>a>>b;100         addedge(a,b);101     }102     F(i,1,n) if(!vis[i]) dfs(i,0);103     F(i,1,n)for(int j=head[i];j!=-1;j=edge[j].next){104         if(num[i]!=num[edge[j].to]){105             du[num[i]]++;106         }107     }108     int x;sum=0;109     F(i,1,cut) 110      if(!du[i]){111             sum++;112             x=i;113     }114     if(sum==1){115         sum=0;116         F(i,1,n) if(num[i]==x) sum++;117         cout<<sum<<endl;118     }119     else cout<<"0"<<endl;120     return 0;121 }
poj 2186
 
 

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