NYOJ 349 Sorting It All Out

Sorting It All Out

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

描述

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.

输入

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.

输出

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. 

样例输入

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

样例输出

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

来源

POJ

上传者

陈玉

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<vector>#include<algorithm>using namespace std;#define s 120#define s 120#define inf 0x3f3f3f3fint n,m,indegree[s],vis[s],tmp[s],rel[s][s];char c[4],ans[s];int toposort(){int i,j,zero,k,flag=1;memcpy(tmp,indegree,sizeof(tmp));memset(ans, '\0', sizeof(ans));      for (i = 0; i < n; i++)      {          zero = 0;//入度为0的点的个数          for (j = 0; j < n; j++){              if (tmp[j] == 0){                  k = j;//下标                  zero++;              }          }          if (zero == 0) return -1;//有环           if (zero>1)  flag = 0;//无序（还不能确定是否有环, 所以继续）          else    ans[i] = 'A' + k;          tmp[k]--;        for (j = 0; j < n; j++){              if (rel[k][j])//由此点出发能直达的点入度减 1                  tmp[j]--;          }      }    return flag;} int main(){int i,j,k;while(~scanf("%d%d",&n,&m)&&n+m){bool sure=false;memset(rel,0,sizeof(rel));memset(indegree,0,sizeof(indegree));for(i=1;i<=m;i++){scanf("%s",&c);if(sure) continue;if(!rel[c[0]-'A'][c[2]-'A']){rel[c[0]-'A'][c[2]-'A']=1;indegree[c[2]-'A']++;}//判重 int res=toposort();if(res==1){sure=true;printf("Sorted sequence determined after %d relations: %s.\n", i, ans);}else if(res==-1){sure=true;printf("Inconsistency found after %d relations.\n", i);}}if(!sure)  printf("Sorted sequence cannot be determined.\n"); }return 0;}