P2341 [HAOI2006] popular cattle | (tarjan)

Topic background

This problem has been fixed test data.

Title Description

Each cow dreams of becoming a star in the bullpen. Like all cows are cows cow is a star. All milk

Cattle are narcissistic per cow always like to own. Between cows "like" can be delivered - if A hi

Huan B, B like C, then A also like C. A total of N cowshed cows given love relationship between the number of cows, you

Calculate how many cows can become a star.

Input Format

 first line: space-separated by two integers: N and M

 second row to row M + 1: an integer of two space-separated each line: A and B, and A represent like B

Output Format

 The first line: a single integer representing the number of stars cows

Sample input and output

Input # 1

Output # 1

Description / Tips

Only 3 cows can be a star

【data range】

10% of the data N <= 20, M <= 50

30% of the data N <= 1000, M <= 20000

70% of the data N <= 5000, M <= 50000

100% data N <= 10000, M <= 50000

Problem-solving ideas: the title is meant to find all the cattle like cattle to reduce the number of points and then find the point of zero degrees

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <vector>
 5 #include <algorithm>
 6 #include <stack>
 7 using namespace std;
 8 int n,m,tot=0,cnt;
 9 const int maxn=5e4+5;
10 vector<int> G[maxn];
11 stack<int> sta;
12 int dfn[maxn],low[maxn],vis[maxn],save[maxn],dis[maxn],out[maxn],x[maxn],y[maxn];
13 
14 void tarjan(int ee){
15     dfn[ee]=low[ee]=++tot;
16     vis[ee]=1;sta.push(ee);
17     for(int i=0;i<G[ee].size();i++){
18         int u=G[ee][i];
19         if(dfn[u]==0){
20             tarjan(u);
21             low[ee]=min(low[ee],low[u]);
22         }else if(vis[u]){
23             low[ee]=min (Low [EE], DFN [U]);
 24          }
 25      }
 26 is      IF (DFN [EE] == Low [EE]) {
 27          CNT ++ ;
 28          the while (! sta.empty ()) {
 29              int Top = sta.top ();
 30              sta.pop ();
 31 is              VIS [Top] = 0 ;
 32              DIS [CNT] ++;       /// this strongly connected components inside how many values 
33 is              Save [Top] = CNT;    / // this node which belongs to the strongly connected component 
34 is              IF (Top == EE) BREAK ;
 35          }
 36     }
37 }
38 
39 int main(){
40     scanf("%d%d",&n,&m);
41     for(int i=1;i<=m;i++){
42         scanf("%d%d",&x[i],&y[i]);
43         G[x[i]].push_back(y[i]);
44     }
45     for(int i=1;i<=n;i++){
46         if(dfn[i]==0) tarjan(i);
47     }
48     for ( int I = . 1 ; I <= m; I ++ ) {
 49          IF (Save [X [I]] = Save [Y [I]]!) OUT [Save [X [I]]] ++;    // / the strongly connected components of a number of 
50      }
 51 is      int in Flag = 0 , RES;
 52 is      for ( int I = . 1 ; I <= cnt; I ++) {    /// number cnt strongly connected component 
53 is          IF ( oUT [ I] == 0 ) {
 54 is              In Flag ++; RES = DIS [I];
 55          }
 56 is      }
 57 is      the printf ( "%d\n",flag==1?res:0);
58     return 0;
59 }

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