UVA 11504 - Dominos

题目链接

题意：给定一个多米诺骨牌的有向图，为最多要推几个才能全倒

思路：强连通分量，缩点后找出度数为0的点就是答案

代码：

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        using namespace std; const int N = 100005; vector
       
         g[N], scc[N]; int pre[N], lowlink[N], sccno[N], dfs_clock, scc_cnt; stack
        
          S; void dfs_scc(int u) { pre[u] = lowlink[u] = ++dfs_clock; S.push(u); for (int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if (!pre[v]) { dfs_scc(v); lowlink[u] = min(lowlink[u], lowlink[v]); } else if (!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]); } if (lowlink[u] == pre[u]) { scc_cnt++; while (1) { int x = S.top(); S.pop(); sccno[x] = scc_cnt; if (x == u) break; } } } void find_scc(int n) { dfs_clock = scc_cnt = 0; memset(sccno, 0, sizeof(sccno)); memset(pre, 0, sizeof(pre)); for (int i = 0; i < n; i++) if (!pre[i]) dfs_scc(i); } int in[N]; int t, n, m, u[N], v[N]; int main() { scanf("%d", &t); while (t--) { scanf("%d%d", &n, &m); for (int i = 0; i < n; i++) g[i].clear(); for (int i = 0; i < m; i++) { scanf("%d%d", &u[i], &v[i]); u[i]--; v[i]--; g[u[i]].push_back(v[i]); } find_scc(n); memset(in, 0, sizeof(in)); for (int i = 0; i < m; i++) { if (sccno[u[i]] != sccno[v[i]]) in[sccno[v[i]]]++; } int ans = 0; for (int i = 1; i <= scc_cnt; i++) if (in[i] == 0) ans++; printf("%d\n", ans); } return 0; }