首先是贪心的思想。。。我们从小到大确定每个最前的位置
建出反图，然后拓扑排序，每次找$deg=0$的点中最大的那个，于是就可以保证编号小的尽可能的在后面
最后把顺序倒过来输出就好了。。。
1 /************************************************************** 2 Problem: 4010 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:900 ms 7 Memory:3936 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <queue> 13 14 using namespace std; 15 const int N = 1e5 + 5; 16 17 struct edge { 18 int next, to; 19 edge(int _n = 0, int _t = 0) : next(_n), to(_t) {} 20 } e[N]; 21 22 int n, m; 23 int first[N], tot; 24 int deg[N], q[N]; 25 int t[N], NOW, v[N]; 26 int ans[N]; 27 priority_queue <int> h; 28 29 inline int read(); 30 31 inline void add_edge(int x, int y) { 32 e[++tot] = edge(first[x], y), first[x] = tot; 33 ++deg[y]; 34 } 35 36 #define p q[l] 37 #define y e[x].to 38 bool check_circle() { 39 static int i, l, r, x; 40 for (r = 0, i = 1; i <= n; ++i) if (deg[i] == 0) q[++r] = i; 41 for (l = 1; l <= r; ++l) 42 for (x = first[p]; x; x = e[x].next) 43 if (--deg[y] == 0) q[++r] = y; 44 return r != n; 45 } 46 47 inline int bfs(int s) { 48 static int l, r, x; 49 ++NOW; 50 q[l = r = 1] = s; 51 for (; l <= r; ++l) 52 for (x = first[p]; x; x = e[x].next) if (!v[y]) { 53 ++deg[y]; 54 if (t[y] != NOW) 55 q[++r] = y, t[y] = NOW; 56 } 57 } 58 #undef p 59 60 inline void get(int s) { 61 static int p, x; 62 while (!h.empty()) h.pop(); 63 h.push(s), v[s] = 1; 64 while (!h.empty()) { 65 p = h.top(), h.pop(); 66 ans[++ans[0]] = p; 67 for (x = first[p]; x; x = e[x].next) if (!v[y]) 68 if (--deg[y] == 0) h.push(y), v[y] = 1; 69 } 70 } 71 #undef y 72 73 int main() { 74 int T, i, x, y; 75 T = read(); 76 while (T--) { 77 n = read(), m = read(), tot = NOW = 0; 78 memset(first, 0, sizeof(first)); 79 memset(v, 0, sizeof(v)); 80 memset(deg, 0, sizeof(deg)); 81 memset(t, 0, sizeof(t)); 82 for (i = 1; i <= m; ++i) { 83 x = read(), y = read(); 84 add_edge(y, x); 85 } 86 if (check_circle()) { 87 puts("Impossible!"); 88 continue; 89 } 90 for (i = 1; i <= n; ++i) if (!v[i]) { 91 bfs(i); 92 get(i); 93 for (; ans[0]; --ans[0]) printf("%d ", ans[ans[0]]); 94 } 95 puts(""); 96 } 97 return 0; 98 } 99 100 inline int read() { 101 static int x; 102 static char ch; 103 x = 0, ch = getchar(); 104 while (ch < '0' || '9' < ch) 105 ch = getchar(); 106 while ('0' <= ch && ch <= '9') { 107 x = x * 10 + ch - '0'; 108 ch = getchar(); 109 } 110 return x; 111 }