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POJ 1094-Sorting It All Out(拓扑排序)
Sorting It All Out
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 28055 | Accepted: 9689 |
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.
每输入一条边就要用拓扑排序判断一次,如果有环则输出有环,如果多解则继续输入,如果排序已完成则输出排序结果。
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <iostream> #include <algorithm> #include <vector> #include <stack> using namespace std; int n, m,in[30],cnt,ma[30][30]; char ans[30]; int top_sort() { stack <int> s; int flag = 0; cnt=0; memset(in,0,sizeof(in)); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(ma[i][j]) in[j]++; for (int i = 1; i <= n; i++) if (in[i] == 0) { s.push(i); } while (!s.empty()) { if (s.size() > 1) { flag = 1; } int u = s.top(); s.pop(); ans[cnt++] = u - 1 + 'A'; for (int i = 1; i <=n; i++) { if(ma[u][i]) { in[i]--; if(in[i]==0) s.push(i); } } } if (cnt < n) { return 0; } else if (flag == 1) { return 1; } else { return 2; } } int main() { char c[5]; while (~scanf("%d%d", &n, &m)) { if (!m && !n) { break; } int ok = 1;memset(ma,0,sizeof(ma)); for (int i = 1; i <= m; i++) { scanf("%s", c); if(!ok)continue; int u = c[2] - 'A' + 1, v = c[0] - 'A' + 1; ma[u][v]=1; int x = top_sort(); if (x == 2) { printf("Sorted sequence determined after %d relations: ", i); for (int i = cnt - 1; i >= 0; i--) { printf("%c", ans[i]); } printf(".\n"); ok = 0; } else if (x == 0) { printf("Inconsistency found after %d relations.\n", i); ok = 0; } } if (ok) { printf("Sorted sequence cannot be determined.\n"); } } return 0; }
POJ 1094-Sorting It All Out(拓扑排序)
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