Sorting It All Out（拓扑排序）

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Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33626 Accepted: 11753

Description

An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.

Input

Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.

Output

For each problem instance, output consists of one line. This line should be one of the following three:

Sorted sequence determined after xxx relations: yyy...y.
Sorted sequence cannot be determined.
Inconsistency found after xxx relations.

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, and yyy...y is the sorted, ascending sequence.

Sample Input

4 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 0

Sample Output

Sorted sequence determined after 4 relations: ABCD.Inconsistency found after 2 relations.Sorted sequence cannot be determined.

Source

East Central North America 2001

这题典型的拓扑排序，但是有点变化。
题目样例的三种输出分别是：
1. 在第x个关系中可以唯一的确定排序，并输出。
2. 在第x个关系中发现了有回环（Inconsisitency矛盾）
3.全部关系都没有发现上面两种情况，输出第3种.


那么对于给定的m个关系，一个个的读进去，每读进去一次就要进行一次拓扑排序，如果发现情况1和情况2，那么就不用再考虑后面的那些关系了，但是还要继续读完后面的关系（但不处理）。如果读完了所有关系，还没有出现情况1和情况2，那么就输出情况3.
拓扑排序有两种方法，一种是算法导论上的，一种是用贪心的思想，这题用贪心的思想做更好。


贪心的做法：
1. 找到所有入度为0的点， 加入队列Q
2.取出队列Q的一个点，把以这个点为起点，所有它的终点的入度都减1. 如果这个过程中发现经过减1后入度变为0的，把这个点加入队列Q。
3.重复步骤2，直到Q为空。


这个过程中，如果同时有多个点的入度为0，说明不能唯一确定关系。
如果结束之后，所得到的经过排序的点少于点的总数，那么说明有回环。




题目还需要注意的一点：如果边(u,v)之前已经输入过了，那么之后这条边都不再加入。

#include <iostream>#include <cstdio>#include <stack>#include <cstdlib>#include <cstring>using namespace std;int n,m;int mp[110][110],in[110],list[110],tm[110];//list数组存储输出的字母，mp[i][j]=1则存在关系且i<jint toopsort(){    stack<int>s;    while(!s.empty())    {       s.pop();    }    int flag=0,j=0;    memcpy(tm,in,sizeof(in));    for(int i=0;i<n;i++)    {        if(tm[i]==0)        {          s.push(i);//找出入度为0得点即为起始点        }    }    while(!s.empty())    {         if(s.size()>1)//如果入度为零得点有很多，则很多点还没涉及到还处于初始化中，为1时才只有一个起始点         {            flag=1;//不能确定关系则要继续进行         }         int t=s.top();         s.pop();         list[j++]=t;         for(int i=0;i<n;i++)         {            if(mp[t][i])//t和i有关系            {               if(--tm[i]==0)//存储入度为0得点               {                  s.push(i);               }            }         }    }    if(j!=n)//如果j>n则说明有环，应为处理完后加上初始化的入度为0得点应为n个    return 1;    else if(flag)//无法确定关系继续    return 2;    else return 0;//既没有环又没有不能确定关系的则完成}int main(){    int flag1,flag2;    char x,y;    while(~scanf("%d%d",&n,&m)&&n&&m)    {          getchar();          flag1=0;          flag2=0;          memset(mp,0,sizeof(mp));          memset(in,0,sizeof(in));          for(int i=1;i<=m;i++)          {          scanf("%c<%c",&x,&y);//即使有了结果也要保证全部输入完成          getchar();          if(!flag1&&!flag2)//用两个标记变量来确定          {               if(mp[y-'A'][x-'A']==1)//xY和Yx都有               {                  printf("Inconsistency found after %d relations.",i);                  flag1=1;                  continue;               }               if(mp[x-'A'][y-'A']==0)               {                  mp[x-'A'][y-'A']=1;                  in[y-'A']++;               }               int tmp=toopsort();               if(tmp==1)               {                  printf("Inconsistency found after %d relations.",i);                  flag1=1;               }               else if(tmp==0)               {                  flag2=1;                  printf("Sorted sequence determined after %d relations: ",i);                  for(int j=0;j<n;j++)                  {                     printf("%c",list[j]+'A');                  }                  printf(".");               }          }          }          if(!flag1&&!flag2)//剩下的情况就是无法判定了          {          printf("Sorted sequence cannot be determined.");          }          printf("\n");    }    return 0;}