HDU 3342 Legal or Not (有向图的拓扑排序)

Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 700    Accepted Submission(s): 295

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
  
  

 

解题思路：

      有向图的拓扑排序。

#include <iostream> using namespace std; #define MaxVertexNum 100 typedef int VertexType; typedef struct node //边表结点 { int adjvex;//邻接点域 struct node* next;//链域 //若要表示边上的权，则应增加一个数据域 } EdgeNode; typedef struct vnode //顶点表结点 { VertexType vertex;//顶点域 EdgeNode* firstedge;//边表头指针 int count; } VertexNode; typedef VertexNode AdjList[MaxVertexNum];//AdjList是邻接表类型 typedef struct { AdjList adjlist;//邻接表 int n,e;//顶点数和边数 } ALGraph; //对于简单的应用，无需定义此类型，可直接使用AdjList类型 bool CreateALGraph ( ALGraph *G ) { int i,j,k; EdgeNode* s; cin>>G->n>>G->e; if ( G->n==0 ) return false; for ( i=0; i<G->n; i++ ) { G->adjlist[i].vertex=i; G->adjlist[i].firstedge=NULL; } for ( k=0; k<G->e; k++ ) { cin>>j>>i; s= ( EdgeNode* ) malloc ( sizeof ( EdgeNode ) ); s->adjvex=j; s->next=G->adjlist[i].firstedge; G->adjlist[i].firstedge=s; } return true; } void TopSort ( ALGraph *G ) { int i,j,k=0; int St[MaxVertexNum],top=-1; EdgeNode *p; for ( i=0; i<G->n; i++ ) G->adjlist[i].count=0; for ( i=0; i<G->n; i++ ) { p=G->adjlist[i].firstedge; while ( p!=NULL ) { G->adjlist[p->adjvex].count++; p=p->next; } } for ( i=0; i<G->n; i++ ) { if ( G->adjlist[i].count==0 ) { top++; St[top]=i; } } while ( top>-1 ) { i=St[top]; top--; k++; //printf("%d ",i); p=G->adjlist[i].firstedge; while ( p!=NULL ) { j=p->adjvex; G->adjlist[j].count--; if ( G->adjlist[j].count==0 ) { top++; St[top]=j; } p=p->next; } } if ( k<G->n ) cout<<"NO"<<endl; else cout<<"YES"<<endl; } int main() { ALGraph G; while ( CreateALGraph ( &G ) ) TopSort ( &G ); return 0; }