POJ 1094 Sorting It All Out

Posted 2020-09-07 谦谦君子，陌上其华

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

题意：

给出n个大写字母和m个关系式，判断是否有序排列并求出排列顺序。

 

思路：
拓扑排序。根据题意的话每读一个关系式都需要进行拓扑排序，检验由此是否可以求出排列顺序或者判断出是否有环。

关于拓扑排序：

每次寻找入度为0的点，从该点出发，删掉有该点有关的边并使与该点相连的点的入度-1，重复上述步骤，直至把所有点都挑出来。如果中国有环的话，那么到时候肯定就没有入度为0的点了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 using namespace std;
 7 
 8 const int maxn = 30;
 9 
10 int n, m;
11 int g[maxn][maxn];
12 int indegree[maxn];
13 int ans[maxn];
14 
15 int tuopu()
16 {
17     int temp[maxn];
18     int flag = 1;
19     memcpy(temp, indegree, sizeof(indegree));
20     int cnt = 0;
21 
22     for (int i = 0; i < n; i++)
23     {
24         int num = 0, loc;
25         for (int j = 0; j < n; j++)
26         {
27             if (temp[j] == 0)
28             {
29                 num++;
30                 loc = j;
31             }
32         }
33         if (num == 0)   return 0;   //没有入度为0的点，说明有环
34         if (num>1)    flag=-1;
35         ans[cnt++] = loc;
36         temp[loc] = -1;
37         for (int j = 0; j < n;j++)
38         if (g[loc][j])   temp[j]--;
39     }
40     return flag;
41 }
42 
43 int main()
44 {
45     //freopen("D:\\txt.txt", "r", stdin);
46     char s[5];
47     while (~scanf("%d%d", &n, &m))
48     {
49         if (m == 0 && n == 0) break;
50         memset(g, 0, sizeof(g));
51         memset(indegree, 0, sizeof(indegree));
52         memset(ans, 0, sizeof(ans));
53         int flag = 0;
54         for (int i = 1; i <= m;i++)
55         {
56             scanf("%s", &s);    
57             if (flag)  continue;
58             int x = s[0] - ‘A‘;
59             int y = s[2] - ‘A‘;
60             g[x][y] = 1;
61             indegree[y]++;      //y的入度+1
62             int p = tuopu();
63             if (p == 0)
64             {
65                 printf("Inconsistency found after %d relations.\n", i);
66                 flag = 1;
67             }
68             if (p == 1)
69             {
70                 printf("Sorted sequence determined after %d relations: ", i);
71                 for (int j = 0; j < n; j++)
72                     printf("%c", ans[j] + ‘A‘);
73                 printf(".\n");
74                 flag = 1;
75             }
76         }
77         if (!flag)
78             printf("Sorted sequence cannot be determined.\n");
79     }
80     return 0;
81 }