[POJ 1094]Sorting It All Out[拓扑排序]

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 35774 Accepted: 12595

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 6
A

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

题意

题目给出N个字母(从A开始的依次N个字母)，M对字母之间的关系。让你判断是否能排好序，或是出现矛盾，或是都不能完成。关键在于理解题意，如果前一部分已经将N个字母排好序了，则后面的即使发生冲突也不用管，需要输出完成排序的关系序号。如果发生冲突，也需要输出冲突的关系序号。

题解

理解好题意之后，题解就比较简单了，就是每输入一对关系，都进行一次拓扑排序，判断一下是否已经完成排序或是发出冲突（即存在环）即可，如果有以上情况，则之后的都无需再判断，如果最后都没有完成排序，输出没有完成即可。

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int N, M;
int mat[30][30], in[30], que[30], tot, vis[30], tmp[30];
void init()
{
  memset(mat, 0, sizeof mat);
  memset(in, 0, sizeof in);
  tot = 0;
}
int topsort()
{
  tot = 0;
  int cnt = 0, ret = 0;
  int src;
  memcpy(tmp, in, sizeof tmp);

  for(int i = 1; i <= N; i++)
  {
    cnt = 0;
    for(int j = 1; j <= N; j++)
    {
      if(tmp[j] == 0){
        cnt++;
        src = j;
      }
    }
    if(cnt == 0)  return -1;
    if(cnt > 1) ret = 1;
    que[tot++] = src;
    tmp[src] = -1;
    for(int j = 1; j <= N; j++)
      if(mat[src][j]) tmp[j]--;
  }
  return ret;
}
int main()
{
  while(~scanf("%d%d", &N, &M))
  {
    init();
    int flag = 1, err, rit;
    if(N == 0 && M == 0)  break;
    for(int i = 1; i <= M; i++)
    {
      char str[10];
      int u, v;
      scanf("%s", str);
      u = str[0] - 'A' + 1, v = str[2] - 'A' + 1;
      mat[u][v] = 1;
      in[v]++;
      if(flag == 1)
      {
        flag = topsort();
        if(flag == -1){
          err = i;
          printf("Inconsistency found after %d relations.\n", err);
        }
        if(flag == 0){
          rit = i;
          printf("Sorted sequence determined after %d relations: ", rit);
          for(int i = 0; i < N; i++)
            printf("%c", que[i] - 1 + 'A');
          printf(".\n");
        }
      }
    }
    if(flag == 1)
      printf("Sorted sequence cannot be determined.\n");
  }
  return 0;
}