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  • POJ——T2186 Popular Cows || 洛谷——P2341 [HAOI2006]受欢迎的牛

    http://poj.org/problem?id=2186

    ||

    https://www.luogu.org/problem/show?pid=2341

    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 33470   Accepted: 13634

    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
    

    Hint

    Cow 3 is the only cow of high popularity. 

    Source

    tarjan缩点,输出出度为0的点的子图中的点的个数

     1 #include <algorithm>
     2 #include <cstring>
     3 #include <cstdio>
     4 
     5 using namespace std;
     6 
     7 const int N(50015);
     8 int n,m,u,v,sumchudu,ans,cnt;
     9 int point[N],chudu[N];
    10 int sumedge,head[N];
    11 struct Edge
    12 {
    13     int from,to,next;
    14     Edge(int from=0,int to=0,int next=0) :
    15     from(from),to(to),next(next) {}
    16 }edge[N];
    17 
    18 int ins(int from,int to)
    19 {
    20     edge[++sumedge]=Edge(from,to,head[from]);
    21     return head[from]=sumedge;
    22 }
    23 
    24 int dfn[N],low[N],tim;
    25 int Stack[N],top,instack[N];
    26 int col[N],sumcol;
    27 
    28 void DFS(int now)
    29 {
    30     dfn[now]=low[now]= ++tim;
    31     Stack[++top]=now; instack[now]=1;
    32     for(int i=head[now];i;i=edge[i].next)
    33     {
    34         v=edge[i].to;
    35         if(instack[v]) low[now]=min(low[now],dfn[v]);
    36         else if(!dfn[v])
    37               DFS(v), low[now]=min(low[now],low[v]);
    38     }
    39     if(low[now]==dfn[now])
    40     {
    41         col[now]= ++sumcol;
    42         point[sumcol]++;
    43         for(;Stack[top]!=now;top--)
    44         {
    45             point[sumcol]++;
    46             col[Stack[top]]=sumcol;
    47             instack[Stack[top]]=0;
    48         }
    49         instack[now]=0; top--;
    50     }
    51 }
    52 
    53 int main()
    54 {
    55     scanf("%d%d",&n,&m);
    56     for(int i=1;i<=m;i++)
    57         scanf("%d%d",&u,&v),ins(u,v);
    58     for(int i=1;i<=n;i++)
    59         if(!dfn[i]) DFS(i);
    60     for(int i=1;i<=m;i++)
    61     {
    62         u=edge[i].from; v=edge[i].to;
    63         if(col[u]!=col[v]) chudu[col[u]]++;
    64     }
    65     for(int i=1;i<=sumcol;i++) if(!chudu[i])
    66         ++sumchudu,ans=point[i];
    67     if(sumchudu!=1) ans=0;
    68     printf("%d
    ",ans);
    69     return 0;
    70 }
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
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  • 原文地址:https://www.cnblogs.com/Shy-key/p/6827923.html
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