POJ 1094 Sorting It All Out

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 31877 Accepted: 11086

Description

An ascending(上升的) sortedsequence(序列) ofdistinct(明显的) values is one in which some form of a less-than operator is used to order theelements(基础) from smallest to largest. For example, the sorted sequence A, B, C, Dimplies(意味) 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(指定) or not.

Input

Input(投入) consists of multiple probleminstances(实例). Each instance starts with a line containing twopositive(积极的)integers(整数) n and m. the first valueindicated(表明) the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of theuppercase(以大写字母印刷)alphabet(字母表). 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(输出) consists of one line. This line should be one of the following three:

Sorted sequence(序列) determined after xxx relations: yyy...y.
Sorted sequence cannot be determined.
Inconsistency(不一致) 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(上升的) 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.

该题题意明确，就是给定一组字母的大小关系判断他们是否能组成唯一的拓扑序列。是典型的拓扑排序，拓扑排序可以用栈来实现，每次入栈的是入度为0的节点。但输出格式上却有三种形式：
1.该字母序列有序，并依次输出；
2.该序列不能判断是否有序；
3.该序列字母次序之间有矛盾，即有环存在。
而这三种形式的判断是有顺序的：先判断是否有环（3），再判断是否有序（1），最后才能判断是否能得出结果（2）。注意：对于（2）必须遍历完整个图，而（1）和（3）一旦得出结果，对后面的输入就不用做处理了。

#include <stdio.h>#include <string.h>int N, M;int map[30][30];int indegree[30];int in[30];int order[30];int TopoSort(){memcpy(in, indegree, sizeof(indegree));int cas = 1, len = 0, pos, num, i, j;for(i = 1; i <= N; i++){num = 0;for(j = 1; j <= N; j++){if(in[j] == 0){num++;pos = j;}}if(num == 0) return 0;if(num > 1) cas = -1;order[len++] = pos;in[pos] = -1;for(j = 1; j <= N; j++)if(map[pos][j] == 1)in[j]--;}return cas;}int main(){int i, j, cas, x, y;bool done;char s[5];while(scanf("%d%d", &N, &M) != EOF && N && M){done = false;memset(map, 0, sizeof(map));memset(indegree, 0, sizeof(indegree));for(i = 1; i <= M; i++){scanf("%s", s);if(done) continue;x = s[0] - 'A' + 1;y = s[2] - 'A' + 1;if(map[y][x] == 1){done = true;printf("Inconsistency found after %d relations.\n", i);continue;}map[x][y] = 1;indegree[y]++;cas = TopoSort();if(cas == 1){printf("Sorted sequence determined after %d relations: ", i);for(j = 0; j < N; j++)printf("%c", order[j] + 'A' - 1);printf(".\n");done = true;}else if(cas == 0){printf("Inconsistency found after %d relations.\n", i);done = true;}}if(!done)printf("Sorted sequence cannot be determined.\n");}return 0;}