zoukankan      html  css  js  c++  java

  • poj.1094.Sorting It All Out（topo)

    Sorting It All Out
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 28762   Accepted: 9964

    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 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
      1 #include<stdio.h>
  2 #include<string.h>
  3 #include<queue>
  4 using namespace std;
  5 int n , m ;
  6 bool map[30][30] ;
  7 int in[30] ;
  8 char st[1000][5] ;
  9 char a[30] ;
 10 
 11 int topo ()
 12 {
 13     queue <int> q ;
 14     int indegree[30] ;
 15     int cnt = 0 ;
 16     int k = 0 ;
 17     int ans = -1 ;//record the condition that "cnt > 1"
 18     while (!q.empty ())
 19         q.pop () ;
 20     for (int i = 1 ; i <= n ; i++)
 21         indegree[i] = in[i] ;
 22     for (int i = 1 ; i <= n ; i++) {
 23         if (in[i] == 0) {
 24             q.push (i) ;
 25             cnt++ ;
 26         }
 27       //  printf ("%d " , in[i]) ;
 28     }
 29     if (cnt > 1)
 30         ans = 1 ;//conditons are not satisfied
 31     while (!q.empty ()) {
 32         int temp = q.front () ;
 33         a[k++] = 'A' + temp - 1 ;
 34         q.pop () ;
 35         cnt = 0 ;
 36         for (int i = 1 ; i <= n ; i++) {
 37             if (map[temp][i]) {
 38                 indegree[i]-- ;
 39                 if (indegree[i] == 0) {
 40                     cnt++ ;
 41                     q.push (i);
 42                 }
 43             }
 44         }
 45         if (cnt > 1)
 46             ans = 1 ;//conditons are not satisfied
 47     }
 48     if (k != n )
 49         return 0 ;//there is a circle
 50     if (k == n && ans != 1)
 51         return  2 ;//success
 52 
 53     if (ans == 1)
 54         return 1 ;
 55 }
 56 
 57 int main ()
 58 {
 59    // freopen ("a.txt" , "r" , stdin) ;
 60     int link ;
 61     int flag ;
 62     int success ;
 63     while (~ scanf ("%d%d" , &n , &m)) {
 64         if (n == 0 && m ==0)
 65             break ;
 66         getchar () ;
 67         memset (in , 0 , sizeof(in)) ;
 68         memset (map , 0 , sizeof(map)) ;
 69         link = -1 ;
 70         flag = -1 ;
 71         success = -1 ;
 72         for (int i = 0 ; i < m ; i++)
 73             gets (st[i]) ;
 74 
 75         for (int i = 0 ; i < m ; i++) {
 76             int u = st[i][0] - 'A' + 1 ;
 77             int v = st[i][2] - 'A' + 1 ;
 78          //   printf ("u = %d , v = %d
" , u , v) ;
 79             if (!map[u][v])
 80                 in[v]++ ;
 81             map[u][v] = 1 ;
 82 
 83             flag = topo () ;
 84             if (flag == 0) {
 85                 link = i + 1 ;
 86                 break ;
 87             }
 88             else if(flag == 2) {
 89                 success = i + 1 ;
 90                 break ;
 91             }
 92          //   puts ("") ;
 93         }
 94         if (flag == 0) {
 95             printf ("Inconsistency found after %d relations.
" , link) ;
 96         }
 97         else if (flag == 1) {
 98             puts ("Sorted sequence cannot be determined.") ;
 99         }
100         else {
101             printf ("Sorted sequence determined after %d relations: " , success) ;
102             for (int i = 0 ; i < n ; i++) {
103                 printf ("%c" , a[i]) ;
104             }
105             printf (".
") ;
106         }
107     }
108     return 0 ;
109 }
    View Code

    要一条条边测下来，not determined 肯定是最后一条边出来才能 判断 ，但在这之前若 先判断出了 success 或 inconsistency 就能结束了（一开始要把所有边先记录下来）

    success只有 无环 & 同一时刻加入队列的点 == 1 时才能 成立
  • 相关阅读:
    [BZOJ] 2054 疯狂的馒头
    day33（sql）
    day32(表单校验js和jquery表单校验)
    day31（正则表达式）
    day30(对象转json(java))
    day29(对象转xml（使用java）)
    day28(ajax之js原生代码实现)
    day27(反射之内省机制实现BeanUtils)
    day27(反射之内省机制)
    day26(分页查询)
  • 原文地址：https://www.cnblogs.com/get-an-AC-everyday/p/4315465.html
Copyright © 2011-2022 走看看