题目链接：

思路：用vector分别保留原图和发图，然后分别从val值为1的点正向遍历，va值为2的点反向遍历，如果某个点这两种方式都可以遍历到，则输出1，否则输出0.

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #define REP(i, a, b) for (int i = (a); i < (b); ++i)#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)using namespace std;const int MAX_N = (100000 + 100);int N, M, val[MAX_N], vis1[MAX_N], vis2[MAX_N];int path[MAX_N];vector
         
           g1[MAX_N], g2[MAX_N];void dfs(int u, int fa){ vis1[u] = 1; REP(i, 0, (int)g1[u].size()) { int v = g1[u][i]; if (!vis1[v] && v != fa && val[v] != 1) dfs(v, u); }}void rdfs(int u, int fa){ vis2[u] = 1; if (val[u] == 1) return; REP(i, 0, (int)g2[u].size()) { int v = g2[u][i]; if (!vis2[v] && v != fa) rdfs(v, u); }}int main(){ cin >> N >> M; FOR(i, 1, N) cin >> val[i]; FOR(i, 1, M) { int u, v; cin >> u >> v; g1[u].push_back(v); g2[v].push_back(u); } FOR(i, 1, N) { if (val[i] == 1) dfs(i, -1); else if (val[i] == 2) rdfs(i, -1); } FOR(i, 1, N) printf("%d\n", (vis1[i] & vis2[i])); return 0;}