hdu 3342 Legal or Not 拓排序

Legal or Not

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

---

拓扑排序，如果可以排序，那就YES，否则NO：

代码：

View Code

 1 #include <iostream>

 2 #include <stdio.h>

 3 #include <string.h>

 4 using namespace std;

 5 

 6 int map[105][105];

 7 int indegree[105];  //每个点的入度

 8 

 9 

10 int Topo(int n)

11 {

12     int i,j,ans;

13     for(i=0;i<n;i++)

14     {

15         ans=n;

16         for(j=0;j<n;j++)

17         {

18             if(indegree[j]==0)        //记下入度为0的点

19             {

20                 indegree[j]--;

21                 ans=j;

22                 break;

23             }

24         }

25         if(ans==n) return 0;

26         for(j=0;j<n;j++)

27         {

28             if(map[ans][j]>0)        //删除入度为0的点时也要对与它相连的每个点的入度减一

29             {

30                 indegree[j]--;

31             }

32         }

33     }

34     return 1;

35 }

36 

37 int main()

38 {

39     int n,m,p,q;

40     int i,j,flag;

41     while(1)

42     {

43         scanf("%d%d",&n,&m);

44         if(n==0 && m==0) break;

45         memset(map,0,sizeof(map));

46         memset(indegree,0,sizeof(indegree));

47         for(i=0;i<m;i++)

48         {

49             scanf("%d%d",&p,&q);

50             if(map[p][q]==0)

51             {

52                 map[p][q]=1;

53                 indegree[q]++;

54             }

55         }

56         flag=Topo(n);

57         if(flag) printf("YES\n");

58         else printf("NO\n");

59     }

60     return 0;

61 }