拓撲排序的核心就是每次找入度為0的點，進入輸出隊列 ，然後將與此點相連的節點入度減1
重復做以上操作。
當做n-1 次後還有點沒進輸出隊列 那麽這些點就是環上的 
因為環上的各點入度都為1 沒有0的 就不能更新。
也就是說拓撲排序一遍之後，如果是DAG所有點都恰好入隊一次
如果有環，那麽一定存在沒有入隊的點。

例題：

Legal or Not
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description:

ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it‘s legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian‘s master and, at the same time, 3xian is HH‘s master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.

Please note that the "master and prentice" relation is transitive. It means that if A is B‘s master ans B is C‘s master, then A is C‘s master.

Input:
The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y‘s master and y is x‘s prentice. The input is terminated by N = 0.

Output:
For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".

Sample Input:
3 2
0 1
1 2
2 2
0 1
1 0
0 0

Sample Output:
YES
NO

題意： 題意：給出一個不一定聯通的圖，判斷圖中是否有環

解：

拓撲排序判環。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<map>
 7 #include<queue>
 8 #include<vector>
 9 #define mod 10000
10 #define inf 336860180
11 #define PI 3.1415926
12 #define ll long long
13 using namespace std;
14 const int N=1e6+19;
15 int n,m,x,y,cnt;//cnt 統計入隊點數 
16 struct node{
17     int u,v,c,ne;
18 }e[N];
19 int h[N],tot,du[N];
20 void add(int u,int v,int c)
21 {
22     du[v]++;
23     tot++;e[tot]=(node){u,v,c,h[u]};h[u]=tot;
24 }
25 queue<int>q;
26 bool tuopu()
27 {
28     for(int i=1;i<=n;++i)
29      if(du[i]==0) q.push(i);
30     while(!q.empty())
31     {
32         int ff=q.front();q.pop();cnt++;
33         for(int i=h[ff];i;i=e[i].ne)
34         {
35             int rr=e[i].v;
36             du[rr]--;
37             if(du[rr]==0) q.push(rr);
38         }
39     }
40     if(cnt==n) return 1;//如果有點沒有更新到，說明有環 
41     return 0;
42 }
43 int main()
44 {
45     while(scanf("%d%d",&n,&m)!=EOF)
46     {
47         if(n==0 && m==0) break;
48         tot=0;cnt=0;
49         for(int i=1;i<=n;++i) h[i]=du[i]=0;
50         for(int i=1;i<=m;++i)
51         {
52             scanf("%d%d",&x,&y);
53             x++,y++;add(x,y,0);
54         }
55         if(tuopu()) puts("YES");
56         else puts("NO");
57     }
58     return 0;
59 }

(?′?‵?)Ｉ L???????

拓撲排序判環