pku 1094

拓扑排序方法如下: 　　
(1)从有向图中选择一个没有前驱(即入度为0)的顶点并且输出它. 　　
(2)从网中删去该顶点,并且删去从该顶点发出的全部有向边. 　　
(3)重复上述两步,直到剩余的网中不再存在没有前驱的顶点为止.
这个题是标准的拓扑排序，但需注意这句话：where xxx is the number of relations processed at

the time either a sorted sequence is determined or an inconsistency is found, whichever

comes first, 也就是说一旦可以确定顺序或者出现环，就要立即输出
鄙人对于拓扑排序算法还不太理解，先奉上源代码（从网上找的），有时间再研究
#include<stdio.h>
 #include<string.h>
 #include<stack>
 using namespace std;
 int g[26][26],degree[26],degreeR[26],path[26];
 int  ring,more;
 int n,m;
 int main(){
    int i,j,k,t;
    char u,v;
    int ele;
    int  sure;
    while(scanf("%d%d",&n,&m),n|m){
        memset(g,0,sizeof(g));
        memset(degreeR,0,sizeof(degreeR));
       sure = 0;
        for(getchar(),i=0;i<m;i++){
           scanf("%c<%c",&u,&v);getchar();          
          g[ u-'A' ][ v-'A' ]=1;
           degreeR[ v-'A' ]++;
            stack<int> _sta;
           for(j=0;j<n;j++) if( degreeR[j] == 0 ) _sta.push(j);

           for(j=0;j<n;j++) degree[j]=degreeR[j];
          more=0;
           ring=0;
           for(j=0;j<n;j++){
              if( _sta.size()>1 ) more=1; 
             if( _sta.empty() ){ ring=1; break; }
              ele = _sta.top(); _sta.pop();
              path[j]=ele;
              for(k=0;k<n;k++){
                  if( g[ele][k]!=0 && --degree[k] == 0 ) {  _sta.push(k); }
               }
          }

          if( j == n && more == 0){
               printf("Sorted sequence determined after %d relations: ",i+1);
               for(j=0;j<n;j++) printf("%c",'A'+path[j]);   printf(".\n");
               sure=true;
               break;
          }
            else if( ring == 1 ){ printf("Inconsistency found after %d relations.\n",i+1);

sure=true; break; }
        }
        if( sure == 0 ) printf("Sorted sequence cannot be determined.\n");
        for(t=i+1;t<m;t++){ scanf("%c<%c",&u,&v); getchar(); }
   }
   return 0;
}