HDU 1827 Summer Holiday

题目链接

题意：中文题

思路：强连通缩点，每个点的权值为强连通中最小值，然后入度为0的点就是答案

代码：

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        using namespace std; const int N = 1005; const int INF = 0x3f3f3f3f; int n, m, val[N]; vector
       
         g[N]; int pre[N], lowlink[N], dfs_clock, sccno[N], scc_cnt, scc_val[N]; stack
        
          S; void dfs(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(v); lowlink[u] = min(lowlink[u], lowlink[v]); } else if (!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]); } if (pre[u] == lowlink[u]) { scc_cnt++; scc_val[scc_cnt] = INF; while (1) { int x = S.top(); S.pop(); scc_val[scc_cnt] = min(scc_val[scc_cnt], val[x]); sccno[x] = scc_cnt; if (x == u) break; } } } void find_scc() { dfs_clock = scc_cnt = 0; memset(pre, 0, sizeof(pre)); memset(sccno, 0, sizeof(sccno)); for (int i = 1; i <= n; i++) if (!pre[i]) dfs(i); } int in[N]; int main() { while (~scanf("%d%d", &n, &m)) { for (int i = 1; i <= n; i++) { g[i].clear(); scanf("%d", &val[i]); } int u, v; while (m--) { scanf("%d%d", &u, &v); g[u].push_back(v); } find_scc(); memset(in, 0, sizeof(in)); for (int u = 1; u <= n; u++) { for (int j = 0; j < g[u].size(); j++) { int v = g[u][j]; if (sccno[u] != sccno[v]) in[sccno[v]]++; } } int ans = 0, cnt = 0; for (int i = 1; i <= scc_cnt; i++) if (!in[i]) { ans += scc_val[i]; cnt++; } printf("%d %d\n", cnt, ans); } return 0; }