拓扑排序判环
拓扑排序的核心就是每次找入度为0的点，进入输出队列 ，然后将与此点相连的节点入度减1
重复做以上操作。
当做n-1 次后还有点没进输出队列 那么这些点就是环上的
因为环上的各点入度都为1 没有0的 就不能更新。
也就是说拓扑排序一遍之后，如果是DAG所有点都恰好入队一次
如果有环，那么一定存在没有入队的点。
例题：
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.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
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
题意： 题意：给出一个不一定联通的图，判断图中是否有环
解：
拓扑排序判环。
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 统计入队点数 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;//如果有点没有更新到，说明有环 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 }
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