POJ 1094 Sorting It All Out （拓扑排序）

POJ 1094 Sorting It All Out （拓扑排序）

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

Sample Output

Sorted sequence determined after 4 relations: ABCD.Inconsistency found after 2 relations.Sorted sequence cannot be determined.

题意

依序给出一些字母之间的大小关系，判断并输出它能否唯一确定一个序列满足这一关系。

思路

依旧是一道拓扑排序的题目。

无法唯一确定一个序列当且仅当在所有关系输入完毕之后，进行拓扑排序的时候存在一个以上入度为0的点。

序列不存在会在图中存在环的情况下出现。

于是，每输入一组关系，进行一次拓扑排序，对此次拓扑排序的结果判断并做出相应的选择。

AC 代码

#include#include#include#include#include#include#includeusing namespace std;#define M 30vectorG[M];int in[M],n,m;char ans[M];int solve(){ int res=1,h[M],top=0; memcpy(h,in,sizeof(h)); //copy 入度数组 queuesk; for(int i=0; i0) //如果入度为0的点同时存在一个以上，说明无法唯一确定序列 res=0; for(int i=0; i<(int)G[p].size(); i++) //消除当前点，临界点入度-1 { int j=G[p][i]; if(--h[j]==0) sk.push(j); } } if(top

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