High-connectivity components of the uva-1 11504 Dominos

High-connectivity components of the uva-1 11504 Dominos

 

 

Q: If a domino card has an edge u-> v, it means that u has fallen v automatically. The minimum number of manual push requests is required.

If some cards belong to the same strongly connected component, if any card is pushed down, all cards are pushed down. After the point can be scaled down through strong connectivity, there must be no card of entry level.

 

Note that this question cannot be used to determine the number of points with the input degree 0, because there may be a number of strongly connected components, but the input degree is not 0. For example

N = 6, m = 6

1-> 2,

2-> 3,

3-> 1,

4-> 5,

5-> 6,

6-> 4.

Although there is no entry point of 0, it needs to be pushed twice.

 

 

#pragma comment(linker, /STACK:10240000,10240000)#include #include 
 
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           #define mod 4294967296#define MAX 0x3f3f3f3f#define lson o<<1, l, m#define rson o<<1|1, m+1, r#define SZ(x) ((int)ans.size())#define MAKE make_pair#define INFL 0x3f3f3f3f3f3f3f3fLL#define mem(a) memset(a, 0, sizeof(a))const double pi = acos(-1.0);const double eps = 1e-9;const int N = 100005;const int M = 20005;typedef long long ll;using namespace std;int n, m;vector 
            
              G[N];int pre[N], low[N], scc[N], dfs_clock, scc_cnt;stack 
             
               S;int T;void dfs(int u) { pre[u] = low[u] = ++dfs_clock; S.push(u); for(int i = 0; i < G[u].size(); i++) { int v = G[u][i]; if(pre[v] == 0) { dfs(v); low[u] = min(low[u], low[v]); } else if(scc[v] == 0) { low[u] = min(low[u], low[v]); } } if(low[u] == pre[u]) { scc_cnt++; for(;;) { int x = S.top(); S.pop(); scc[x] = scc_cnt; if(x == u) break; } }}void find_scc() { dfs_clock = scc_cnt = 0; mem(scc); mem(pre); for(int i = 0; i < n; i++) { if(pre[i] == 0) dfs(i); }}int vis[N];int main() { //freopen(in.txt,r,stdin); cin >> T; while(T--) { cin >> n >> m; for(int i = 0; i < n; i++) G[i].clear(); for(int i = 0; i < m; i++) { int x, y; scanf(%d%d, &x, &y); x--, y--; G[x].push_back(y); } find_scc(); mem(vis); for(int u = 0; u < n; u++) { for(int i = 0; i < G[u].size(); i++) { int v = G[u][i]; if(scc[v] != scc[u]) { vis[ scc[v] ]++; } } } int cnt = 0; //cout << scc_cnt << endl; for(int i = 1; i <= scc_cnt; i++) { if(vis[i] == 0) cnt++; } cout << cnt << endl; } return 0;}
             
            
          
         
        
       
      
     
    
   
  
 

 

 

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