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.
题目大意:
给你n个字母和m个约束条件,让你给出最后能否找到一个序列符合。
解题思路:
拓扑排序。
这道题目= =非常的坑,比如如果你在前面以及确定了一个唯一的序列能够符合约束,后面再给出一个矛盾的条件是无效的。。。
如果排除这些之后,这道题目还是非常简单的。直接对于每次约束进行拓扑排序就可以了。
代码:
#include <queue> #include <cstdio> #include <vector> #include <cstring> #include <algorithm> #include <functional> using namespace std; #define pb push_back const int maxn = 30; vector<int> ans, tot, vec[maxn]; int in[maxn], vis[maxn], tmp[maxn], cnt; int toposort(int n){ priority_queue<int, vector<int>, greater<int> > q; while(!q.empty()) q.pop(); for(int i = 0; i < maxn; ++i) tmp[i] = in[i]; int rec = 0; for(int i = 0; i < 26; ++i) if(!tmp[i] && vis[i]) { q.push(i); ++rec; } if(rec == 1) rec = 0; while(!q.empty()){ int p = q.top(); tot.pb(p); q.pop(); int v, len = vec[p].size(); for(int i = 0; i < len; ++i){ v = vec[p][i]; --tmp[v]; if(!tmp[v]) q.push(v), ++rec; } if(rec == 1) rec = 0; } if( !rec && tot.size() == n) return 1; else if( tot.size() == cnt ) return 0; else return -1; } int main() { // freopen("test.in", "r+", stdin); // freopen("test.out", "w+", stdout); char a, b; int n, m, x, y, res, flag; while(~scanf("%d%d", &n, &m)){ if(n == 0 && m == 0) break; cnt = 0; flag = 0; memset(in, 0, sizeof(in)); memset(vis, 0, sizeof(vis)); for(int i = 0; i < maxn; ++i) vec[i].clear(); for(int i = 1; i <= m; ++i) { scanf(" %c<%c", &a, &b); x = a - 'A'; y = b - 'A'; ++in[y]; if(!vis[x]) ++cnt, vis[x] = 1; if(!vis[y]) ++cnt, vis[y] = 1; vec[x].pb(y); if(flag == 0){ tot.clear(); int k = toposort(n); if(k == 1) { ans = tot; flag = 1; res = i; }else if(k == -1){ flag = -1; res = i; } } } if(flag == 1){ printf("Sorted sequence determined after %d relations: ", res); for(int i = 0; i < ans.size(); ++i){ printf("%c", 'A'+ans[i]); } puts("."); }else if(flag == 0){ printf("Sorted sequence cannot be determined. "); }else{ printf("Inconsistency found after %d relations. ", res); } } return 0; }