POJ 1094 -- Sorting It All Out （拓扑排序）

Sorting It All Out

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1094

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.

 题意：根据给出的关系式判断大小关系，当出现环时输出在第几个关系式出现环，

             当可排序时输出在第几个关系式可排序并输出排序顺序。

             当所有关系式输入后无法排序则输出无法确定排序。

 思路：因为要输出第几个关系式，所以每输入一个关系式都要对图进行拓扑排序，

            若发现环则输出含有环，若可排序则输出可排序，重点在于 当出现多个0

            前驱节点时，应将所有0前驱标记，继续对图排序，看看是否含有环！！ 

            输出遵照以下规则：优先级：输出环，输出完整顺序，输出无法排序。

代码如下：

法1：

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <vector>#include <algorithm>using namespace std;#define s 120#define inf 0x3f3f3f3fint n, m, indegree[s], vis[s], tmp[s];char c[4], ans[s];vector<vector<int>> v;//相当于二维数组int toposort(int obj){int num, flag = 1, k, p = 0, i;for (i = 0; i < n; i++)  tmp[i] = indegree[i];//临时存放入度数组,以免影响下次结果while (obj--){num = 0;//每次重新计数for (i = 0; i < n; i++){if (tmp[i] == 0){k = i, num++;}}if (num > 0){if (num > 1)  flag = 0;                        else{   ans[p++] = k + 'A';   ans[p] = '\0';                        }for (i = 0; i < v[k].size(); i++){tmp[v[k][i]]--;}tmp[k]--;}elsereturn -1;//数据矛盾}if (flag) return p;//返回能排序的位数return 0;//排序不确定}int main(){#ifdef OFFLINEfreopen("t.txt", "r", stdin);#endifint i, j, k;while (~scanf("%d %d", &n, &m) && n&&m){v.clear();   v.resize(n);memset(vis, 0, sizeof(vis));memset(indegree, 0, sizeof(indegree));int obj = 0, det = 0;for (i = 1; i <= m; i++){scanf("%s", c);indegree[c[2] - 'A']++;//入度v[c[0] - 'A'].push_back(c[2] - 'A');//建边if (!vis[c[0] - 'A']){vis[c[0] - 'A'] = 1;obj++;//对象数}if (!vis[c[2] - 'A']){vis[c[2] - 'A'] = 1;obj++;}if (det == 0)//排序不确定才继续{int res = toposort(obj);if (res == -1){k = i, det = -1;//k记录数据组数}else if (res == n){k = i, det = 1;}}}if (det == -1)printf("Inconsistency found after %d relations.\n", k);else if (det == 0)printf("Sorted sequence cannot be determined.\n");elseprintf("Sorted sequence determined after %d relations: %s.\n", k, ans);}return 0;}

法二（与法一差不多，好理解点）

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <vector>#include <algorithm>using namespace std;#define s 120#define inf 0x3f3f3f3fint n, m, indegree[s], vis[s], tmp[s], rel[s][s];char c[4], ans[s];int toposort(){int i, j, zero, k, flag = 1;//surememcpy(tmp, indegree, sizeof(tmp));//将长度为数组tmp长度的indegree数组数据复制到tmp数组memset(ans, '\0', sizeof(ans));for (i = 0; i < n; i++){zero = 0;//入度为0的点的个数for (j = 0; j < n; j++){if (tmp[j] == 0){k = j;//下标zero++;}}if (zero == 0) return -1;//有环 if (zero>1)  flag = 0;//无序（还不能确定是否有环, 所以继续）else    ans[i] = 'A' + k;tmp[k]--;for (j = 0; j < n; j++){if (rel[k][j])//由此点出发能直达的点入度减 1tmp[j]--;}}return flag;}int main(){#ifdef OFFLINEfreopen("t.txt", "r", stdin);#endifint i, j, k;while (~scanf("%d %d", &n, &m) && n + m){bool sure = false;memset(rel, 0, sizeof(rel));memset(indegree, 0, sizeof(indegree));for (i = 1; i <= m; i++){scanf("%s", c);if (sure) continue;if (!rel[c[0] - 'A'][c[2] - 'A']){//判重,避免加入重复边rel[c[0] - 'A'][c[2] - 'A'] = 1;indegree[c[2] - 'A']++;}int res = toposort();if (res == 1){sure = true;printf("Sorted sequence determined after %d relations: %s.\n", i, ans);}else if (res == -1){sure = true;printf("Inconsistency found after %d relations.\n", i);}}if (!sure)printf("Sorted sequence cannot be determined.\n");}return 0;}