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Sorting It All Out
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. 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 |
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解题思路：这题是一个拓扑排序问题，不过过程比较复杂，要处理的情况也比较多！首先题目条件和要求要很清楚，这样分析问题思路就会清晰！
解题代码：
1 #include <stdio.h>
2 #include <string.h>
3 #include <iostream>
4 using namespace std;
5 struct Node{ //与之相连的节点结构体
6 int to;
7 struct Node *next;
8 };
9 Node *Link[27]; //链表保存连接关系
10 Node *tm_node;
11 bool vis[27];
12 int count[27], tm_count[27]; //保存节点入度数据，前者为备份数据，后者为运算数据
13 int n, m, num;
14 char deal[3];
15 const int A = ‘A‘;
16 int result[27], pos; //保存结果
17
18 int TopSort(){
19 int i, j, cnt;
20 bool flag = false;
21 pos = cnt = 0;// cnt为入度为0的节点数，pos为保存结果的下标
22 j = -1; //入度为0的字母
23 for (i = 0; i < n; i ++) //寻找入度为0的位置
24 if(tm_count[i] == 0 && vis[i]){
25 cnt ++;
26 j = i;
27 }
28 while (~j){
29 if(cnt > 1) //如果cnt > 1 则表示有多个入度为0的节点，也就是说可能无法排序
30 flag = true; //不直接return是因为还有可能图中有环，导致数据矛盾
31 for (tm_node = Link[j]; tm_node != NULL; tm_node = tm_node ->next)
32 tm_count[tm_node ->to] --;
33 result[pos ++] = j;
34 tm_count[j] --;
35 j = -1;
36 cnt = 0;
37 for (i = 0; i < n; i ++)
38 if(tm_count[i] == 0 && vis[i]){
39 cnt ++;
40 j = i;
41 }
42 }
43 if(!flag && pos != num) // 图中有环，数据矛盾
44 return -1;
45 if(!flag && pos == num) //图中为一个有序序列，是否是最终结果还需进一步判断
46 return 0;
47 if(flag && pos == num) //图中有多个有序序列，还需进一步判断
48 return 1;
49 if(flag && pos != num) //图中有环，数据矛盾
50 return -1;
51 }
52
53 void init(){
54 num = 0;
55 memset(vis, 0, sizeof(vis));
56 memset(Link, 0, sizeof(Link));
57 memset(count, 0, sizeof(count));
58 memset(tm_count, 0, sizeof(tm_count));
59 }
60
61 void input(){
62 int d1, d2;
63 scanf("%s", deal);
64 d1 = deal[0] - A;
65 d2 = deal[2] - A;
66 tm_node = new Node;
67 tm_node ->next = NULL;
68 tm_node ->to = d2;
69 count[d2] ++;
70 if(Link[d1] == NULL)
71 Link[d1] = tm_node;
72 else{
73 tm_node ->next = Link[d1];
74 Link[d1] = tm_node;
75 }
76 if(vis[d1] == 0){
77 num ++;
78 vis[d1] = true;
79 }
80 if(vis[d2] == 0){
81 num ++;
82 vis[d2] = true;
83 }
84 }
85
86 int main(){
87 int i, j;
88 int value;
89 bool had_ans;
90 while(~scanf("%d%d", &n, &m) && (n && m)){
91 init(); //初始化
92 had_ans = false; //是否已经得出结果
93 for(i = 1; i <= m; i ++){
94 if(had_ans){ //如果已有结果，则无须处理后续数据
95 scanf("%s", deal);
96 continue;
97 }
98 input(); //数据处理
99 for(j = 0; j < n; j ++)
100 tm_count[j] = count[j];
101 value = TopSort(); //拓扑排序
102 if (value == -1){
103 printf ("Inconsistency found after %d relations.\n", i);
104 had_ans = true;
105 }
106 else if(value == 0 && num == n){ //数据量以满足要求，且能排序
107 printf ("Sorted sequence determined after %d relations: ", i);
108 for (j = 0; j < n; j ++)
109 printf ("%c", result[j] + A);
110 printf(".\n");
111 had_ans = true;
112 }
113 }
114 if (value == 1 || (value == 0 && n != num)) // 无法排序
115 printf("Sorted sequence cannot be determined.\n");
116 }
117 return 0;
118 }
poj 1094 / zoj 1060 Sorting It All Out