POJ-1094-Sorting It All Out

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33871 Accepted: 11858

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

这道拓扑排序题有两点需要注意的是，它是没输入一对比较后都要判断一次它是否可以排序或则是否矛盾，如果找到一个矛盾的或则成立的拓扑序列，则对后续输入不作处理，如果到输入的末尾都没找到上述两种情况则输出Sorted sequence cannot be determined.
对于判断部分，要注意判断如果队列中如果元素大于1
就一定无序，因为有序是一对一的，每次度为一的点只可能有一个

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <queue>using namespace std;int Map[30][30],in[30],in1[30],out[30],out1[30],vist[30];int n,m,NUM1,NUM2,NUM3,num;int flag1,flag2,flag3;int str[30],cnt;void pan(int Num){if(flag3 || flag1)//如果有序或冲突则不做后续判断return;cnt = 0;memset(str,0,sizeof(str));queue<int>q;flag2 = 0;memset(vist,0,sizeof(vist));for(int i = 1; i <= n; i++){in1[i] = in[i];out1[i] = out[i];if(in1[i] == 0){vist[i] = 1;q.push(i);}}int pos;while(!q.empty()){int Q = q.size();if(Q>1)flag2 = 1;pos = q.front();if(!flag1)str[cnt++] = pos;q.pop();for(int j = 1; j <= n; j++){if(!vist[j] && Map[pos][j]){in1[j]--;out1[pos]--;if(in1[j] == 0){vist[j] = 1;q.push(j);}}}}for(int i = 1; i <= n; i++){if(in1[i]){flag3 = 1;NUM3 = Num;break;}}if(!flag2 && !flag3 && !flag1){flag1 = 1;NUM1 = Num;}}int main(){while(~scanf("%d %d%*c",&n,&m) && (n||m)){memset(Map,0,sizeof(Map));memset(in,0,sizeof(in));memset(out,0,sizeof(out));num = 0;flag1 = flag3 = 0;while(m--){num++;char A,B;scanf("%c%*c%c%*c",&A,&B);int x = A-'A'+1;int y = B-'A'+1;if(!Map[x][y]){Map[x][y] = 1;in[y]++;out[x]++;}pan(num);}if(flag3)cout<<"Inconsistency found after "<<NUM3<<" relations."<<endl;else if(flag1){cout<<"Sorted sequence determined after "<<NUM1<<" relations: ";for(int i = 0; i < cnt; i++)printf("%c",'A'+str[i]-1);cout<<"."<<endl;}elsecout<<"Sorted sequence cannot be determined."<<endl;}return 0;}