poj 1094 Sorting It All Out (拓扑排序)

http://poj.org/problem?id=1094

Sorting It All Out
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 24505   Accepted: 8487

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
 
【题解】：
　　拓扑排序到底就行，注意：
　　　　当遇到两个入度为0的情况，需要判断是否出现环
　　唉，不说了，代码写得真挫，乱七八糟，不过能AC
　　　
【code】：
  1 /**
  2 Judge Status:Accepted    Memory:704K
  3 Time:0MS     Language:G++
  4 Code Lenght:2296B   Author:cj
  5 */
  6 
  7 #include<iostream>
  8 #include<stdio.h>
  9 #include<string.h>
 10 #include<math.h>
 11 #include<algorithm>
 12 
 13 #define N 30
 14 #define M N*N
 15 using namespace std;
 16 
 17 int map[N][N],mark[N],in[N],tin[N];
 18 char anstr[N];
 19 int n,flag;
 20 
 21 void init()
 22 {
 23     memset(map,0,sizeof(map));
 24     memset(mark,0,sizeof(mark));
 25     memset(in,0,sizeof(in));
 26     memset(tin,0,sizeof(tin));
 27 }
 28 
 29 int topCheck(int id)  //拓扑排序
 30 {
 31     int i,cnt=0,pos=-1;
 32     for(i=0;i<26;i++)
 33     {
 34         if(mark[i]&&!in[i])
 35         {
 36             cnt++;
 37             pos=i;
 38         }
 39     }
 40     if(cnt>1)  //当有多个入度为0的点
 41     {
 42         flag = 1;  //标记出现多个入度为0的点的情况
 43         in[pos] = -1;
 44         anstr[id] = pos+'A';
 45         for(i=0;i<26;i++)
 46         {
 47             if(map[pos][i])     in[i]--;
 48         }
 49         return topCheck(id+1);
 50     }
 51     if(cnt==0)
 52     {
 53         for(i=0;i<26;i++)
 54         {
 55             if(mark[i]&&in[i]>0)  
 56             {
 57                 return 1;   //入度都大于0，出现了环
 58             }
 59         }
 60         if(id!=n||flag)   return 0;  //当排序到入度为0时，并且一直没有出现过多个入度为0的情况，则排序可以完成
 61         anstr[id]='\0';
 62         return 2;
 63     }
 64     in[pos] = -1;
 65     anstr[id] = pos+'A';
 66     for(i=0;i<26;i++)
 67     {
 68         if(map[pos][i])     in[i]--;
 69     }
 70     return topCheck(id+1);  //进入下一层拓扑排序
 71     return 0;
 72 }
 73 
 74 int main()
 75 {
 76     int m;
 77     while(~scanf("%d%d",&n,&m))
 78     {
 79         if(!n&&!m)  break;
 80         char xx[5];
 81         int i;
 82         init();
 83         int temp=0,pos=0;
 84         for(i=1;i<=m;i++)
 85         {
 86             scanf("%s",xx);
 87             int x = xx[0]-'A';
 88             int y = xx[2]-'A';
 89             mark[x] = mark[y] = 1;
 90             tin[y]++;
 91             map[x][y] = 1;
 92             memcpy(in,tin,sizeof(int)*26);
 93             if(temp!=0)  continue;
 94             flag = 0;
 95             temp = topCheck(0);
 96             pos = i;
 97         }
 98         if(temp==2)
 99         {
100             anstr[pos+i]='\0';
101             printf("Sorted sequence determined after %d relations: %s.\n",pos,anstr);
102             continue;
103         }
104         if(temp==1)
105         {
106             printf("Inconsistency found after %d relations.\n",pos);
107             continue;
108         }
109         puts("Sorted sequence cannot be determined.");
110     }
111     return 0;
112 }

 

posted @ 2013-08-19 20:52  crazy_apple  阅读(230)  评论(0)  编辑  收藏  举报