Sorting It All Out
Problem 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.
Example 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
Example Output
Sorted sequence determined after 4 relations: ABCD. Inconsistency found after 2 relations. Sorted sequence cannot be determined.
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.
题目大意:
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.
拓扑排序用贪心的思想:
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 <bits/stdc++.h> using namespace std; const int MAXN=1e2; int N, M, sum; int in[MAXN], step[MAXN];///in数组记录入度,step记录路径 vector<int>z[MAXN];///用vector模拟邻接表存数据 void start() { memset(in,0,sizeof(in)); for(int i=0; i<=N; i++) z[i].clear(); } bool Find(int a, int b) { for(int i=0; i<z[a].size(); i++) { if(z[a][i]==b) return true; } return false; } int zhao() { queue<int>Q; sum=0; for(int i=0; i<N; i++) { if(in[i]==0) Q.push(i); } int fly=false; while(!Q.empty()) { if(Q.size()>1)fly=true;///如果只存在一个入度为0的点,说明存在拓扑序列,反之则不构成 int m=Q.front(); Q.pop(); step[sum++]=m; for(int i=0; i<z[m].size(); i++) { in[z[m][i]]--; if(in[z[m][i]]==0)///借助队列降低时间复杂度 Q.push(z[m][i]); } } if(sum<N)///存在回路 return 2; if(fly)return 3; return 1; } int main() { while(cin>>N>>M&&N&&M) { start(); char aa, bb, cc; int flag=3; int stop; for(int t=1; t<=M; t++) { cin>>aa>>bb>>cc; if(flag!=3)continue; int a=aa-'A', c=cc-'A'; if(!Find(a,c))///判断之前是否已经存过此边 { z[a].push_back(c); in[c]++; } int temp[MAXN]; memcpy(temp,in,sizeof(in)); flag=zhao(); memcpy(in,temp,sizeof(temp));///还原in数组 if(flag!=3) stop=t; } if(flag==1) { cout<<"Sorted sequence determined after "<<stop<<" relations: "; for(int i=0; i<sum; i++) { char p=step[i]+'A'; cout<<p; } cout<<'.'<<endl; } else if(flag==2) cout<<"Inconsistency found after "<<stop<<" relations."<<endl; else cout<<"Sorted sequence cannot be determined."<<endl; } return 0; }