Sorting It All Out
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 21865 | Accepted: 7529 |
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.
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<B A<C B<C C<D B<D A<B 3 2 A<B B<A 26 1 A<Z 0 0
Sample Output
Sorted sequence determined after 4 relations: ABCD. Inconsistency found after 2 relations. Sorted sequence cannot be determined.
/* 这道题WA了好久,其中有几个需要注意的地方 1、当出现正好存在一种情况能够排序完所有节点时,不管以后的边会出现什么情况,都输出 能够排序成功 2、当中间的拓扑排序过程中出现多个几点的入度为0时,只记录当时的状态 (亦即该测试数据要么出现环,要么就是有多组解),不能立即返回, 要继续读边,直到能够排序完成(此时输出有多解的情况)或者出现环。 */ #include<cstdio> #include<iostream> #include<cstring> using namespace std; bool G[30][30]; int d[30]; char s[30]; int toposort(int n) { int num,k,i,j,t; int td[30]; bool flag=true; for(i=1;i<=n;++i) td[i]=d[i]; memset(s,0,sizeof(0)); for(j=0;j<n;++j) { num=0; for(i=1;i<=n;++i) { if(td[i]==0) { k=i; ++num; } } if(num==0) //有环 return -1; if(num>1) //有多种情况,还需继续读边判断 { flag=false; } s[j]='A'+k-1; td[k]=-1; for(t=1;t<=n;++t) { if(G[k][t]) --td[t]; } } s[n]='\0'; if(flag==false) //情况不唯一 return 0; return 1; //全部排好序了返回1. } int main() { int n,m,i,ans,k; bool flag; char temp[5]; while(scanf("%d%d",&n,&m),m||n) { memset(G,false,sizeof(G)); memset(d,0,sizeof(d)); flag=true; for(i=1;i<=m;++i) { scanf("%s",temp); if(!flag) //已经排好序或者有环 continue; int u=temp[0]-'A'+1; int v=temp[2]-'A'+1; if(!G[u][v]) { ++d[v]; G[u][v]=true; } ans=toposort(n); if(ans==1||ans==-1) { k=i; flag=false; } } if(ans==1) { printf("Sorted sequence determined after %d relations: %s.\n",k,s); } else if(ans==-1) { printf("Inconsistency found after %d relations.\n", k); } if(flag&&i>m) printf("Sorted sequence cannot be determined.\n"); } return 0; }