题目链接：http://hihocoder.com/problemset/problem/1174
题意：判断一个有向图是否有环，用拓扑排序，结论就是每次取出点的时候统计一下现在剩下几个点，最后没有剩下点就是无环的。
1 /*
2 ━━━━━┒ギリギリ♂ eye！
3 ┓┏┓┏┓┃キリキリ♂ mind！
4 ┛┗┛┗┛┃＼○／
5 ┓┏┓┏┓┃ /
6 ┛┗┛┗┛┃ノ)
7 ┓┏┓┏┓┃
8 ┛┗┛┗┛┃
9 ┓┏┓┏┓┃
10 ┛┗┛┗┛┃
11 ┓┏┓┏┓┃
12 ┛┗┛┗┛┃
13 ┓┏┓┏┓┃
14 ┃┃┃┃┃┃
15 ┻┻┻┻┻┻
16 */
17 #include <algorithm>
18 #include <iostream>
19 #include <iomanip>
20 #include <cstring>
21 #include <climits>
22 #include <complex>
23 #include <fstream>
24 #include <cassert>
25 #include <cstdio>
26 #include <bitset>
27 #include <vector>
28 #include <deque>
29 #include <queue>
30 #include <stack>
31 #include <ctime>
32 #include <set>
33 #include <map>
34 #include <cmath>
35 using namespace std;
36 #define fr first
37 #define sc second
38 #define cl clear
39 #define BUG puts("here!!!")
40 #define W(a) while(a--)
41 #define pb(a) push_back(a)
42 #define Rint(a) scanf("%d", &a)
43 #define Rll(a) scanf("%lld", &a)
44 #define Rs(a) scanf("%s", a)
45 #define Cin(a) cin >> a
46 #define FRead() freopen("in", "r", stdin)
47 #define FWrite() freopen("out", "w", stdout)
48 #define Rep(i, len) for(int i = 0; i < (len); i++)
49 #define For(i, a, len) for(int i = (a); i < (len); i++)
50 #define Cls(a) memset((a), 0, sizeof(a))
51 #define Clr(a, x) memset((a), (x), sizeof(a))
52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
53 #define lrt rt << 1
54 #define rrt rt << 1 | 1
55 #define pi 3.14159265359
56 #define RT return
57 #define lowbit(x) x & (-x)
58 #define onenum(x) __builtin_popcount(x)
59 typedef long long LL;
60 typedef long double LD;
61 typedef unsigned long long ULL;
62 typedef pair<int, int> pii;
63 typedef pair<string, int> psi;
64 typedef pair<LL, LL> pll;
65 typedef map<string, int> msi;
66 typedef vector<int> vi;
67 typedef vector<LL> vl;
68 typedef vector<vl> vvl;
69 typedef vector<bool> vb;
70
71 const int maxn = 100100;
72 const int maxm = 500500;
73 typedef struct Edge {
74 int u, v;
75 int next;
76 }Edge;
77 Edge edge[maxm];
78 int ecnt, head[maxn];
79 int n, m;
80 int dig[maxn];
81 priority_queue<int, vector<int>, greater<int> > pq;
82
83 void adde(int u, int v) {
84 edge[ecnt].u = u;
85 edge[ecnt].v = v;
86 edge[ecnt].next = head[u];
87 head[u] = ecnt++;
88 }
89
90 int main() {
91 // FRead();
92 int T, u, v;
93 Rint(T);
94 W(T) {
95 ecnt = 0; Clr(head, -1); Cls(dig);
96 Rint(n); Rint(m);
97 Rep(i, m) {
98 Rint(u); Rint(v);
99 adde(u, v);
100 dig[v]++;
101 }
102 while(!pq.empty()) pq.pop();
103 For(i, 1, n+1) if(dig[i] == 0) pq.push(i);
104 int cnt = n;
105 while(!pq.empty()) {
106 u = pq.top(); pq.pop();
107 cnt--;
108 for(int i = head[u]; ~i; i=edge[i].next) {
109 v = edge[i].v;
110 if(--dig[v] == 0) pq.push(v);
111 }
112 }
113 if(cnt != 0) printf("Wrong\n");
114 else printf("Correct\n");
115 }
116 RT 0;
117 }
来源：https://www.cnblogs.com/kirai/p/5580300.html