描述
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.
輸入
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.
輸出
For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".
樣例輸入
3 2
0 1
1 2
2 2
0 1
1 0
0 0
樣例輸出
YES
NO
題意
給你N個人編號0-N-1，M對關系，（a，b），a是b的師傅，判斷所以關系是否合法
題解
一道簡單的拓撲排序題，如果N的人存在拓撲排序輸出YES，否則輸出NO
拓撲排序
每次找一個入度為0的點（u），刪去所有G[u][v]=1（u，v）的邊，In[v]--
如果所有N個點都找過，輸出YES
如果某個點沒有找過，輸出NO
代碼
1 #include<cstdio> 2 int main() 3 { 4 int n,m,a,b; 5 while(scanf("%d%d",&n,&m)!=EOF,n) 6 { 7 int In[105]={0},G[105][105]={0}; 8 for(int i=0;i<m;i++) 9 { 10 scanf("%d%d",&a,&b); 11 if(G[a][b]==0)//避免重復 12 { 13 In[b]++; 14 G[a][b]=1; 15 } 16 } 17 int cnt=0; 18 while(cnt<n) 19 { 20 int p,flag=0; 21 for(int i=0;i<n;i++) 22 { 23 if(In[i]==0) 24 { 25 p=i; 26 flag=1; 27 In[i]=-1; 28 break; 29 } 30 } 31 if(++cnt==n)//所有n個點都找過 32 { 33 printf("YES\n"); 34 break; 35 } 36 if(flag==0)//如果某個點不行 37 { 38 printf("NO\n"); 39 break; 40 } 41 for(int i=0;i<n;i++) 42 { 43 if(G[p][i]==1)//刪去邊 44 { 45 In[i]--; 46 G[p][i]=0; 47 } 48 } 49 } 50 } 51 return 0; 52 }
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