POJ 1094 Sorting It All Out

Sorting It All Out

Time Limit: 1000ms

Memory Limit: 10000KB

This problem will be judged on  PKU. Original ID: 1094
64-bit integer IO format: %lld      Java class name: Main

 

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 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.

Source

East Central North America 2001

 

解题：这题说的是边读入一些关系。如果当前出现矛盾的（有环），就记录出现矛盾的是第几组关系。如果当前组已经能够确定n个字母的关系，也要记录当前是第几组。如果当前关系不唯一。那么继续了。

 

判关系是唯一就是时刻判断队列中的元素是不是大于1.如果大于1，说明同时有几个点入度为0.排序就不确定了。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 105;

18 int n,m,cnt,in[maxn],tmp[maxn],result[maxn];

19 vector<int>g[maxn];

20 queue<int>q;

21 void init() {

22     for(int i = 0; i < maxn; i++) {

23         in[i] = 0;

24         g[i].clear();

25     }

26 }

27 bool Find(int x,int y) {

28     for(int i = 0; i < g[x].size(); i++)

29         if(g[x][i] == y) return false;

30     return true;

31 }

32 int topSort() {

33     while(!q.empty()) q.pop();

34     for(int i = 0; i < n; i++)

35         if(!in[i]) q.push(i);

36     cnt = 0;

37     bool uncertain = false;

38     while(!q.empty()) {

39         int u = q.front();

40         if(q.size() > 1) uncertain = true;

41         q.pop();

42         result[cnt++] = u;

43         for(int i = 0; i < g[u].size(); i++) {

44             if(--in[g[u][i]] == 0) q.push(g[u][i]);

45         }

46     }

47     if(cnt < n) return 1;

48     if(uncertain) return 2;

49     return 3;

50 }

51 int main() {

52     bool ok;

53     char u,v,op;

54     int endPoint,flag;

55     while(scanf("%d %d",&n,&m),n||m) {

56         init();

57         ok = false;

58         getchar();

59         flag = 2;

60         for(int i = 1; i <= m; i++) {

61             scanf("%c%c%c%*c",&u,&op,&v);

62             if(ok) continue;

63             if(op == '<' && Find(v-'A',u-'A')) {

64                 g[v-'A'].push_back(u-'A');

65                 in[u-'A']++;

66             } else if(op == '>' && Find(u-'A',v-'A')) {

67                 g[u-'A'].push_back(v-'A');

68                 in[v-'A']++;

69             }

70             memcpy(tmp,in,sizeof(in));

71             flag = topSort();

72             if(flag != 2) {

73                 endPoint = i;

74                 ok = true;

75             }

76             memcpy(in,tmp,sizeof(in));

77         }

78         if(flag == 3) {

79             printf("Sorted sequence determined after %d relations: ", endPoint);

80             for(int i = cnt-1; i >= 0; i--)

81                 printf("%c",result[i]+'A');

82             puts(".");

83         } else if(flag == 1) printf("Inconsistency found after %d relations.\n",endPoint);

84         else puts("Sorted sequence cannot be determined.");

85     }

86     return 0;

87 }

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