POJ1094 Sorting It All Out
Sorting It All Out
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 40012 | Accepted: 14072 |
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.
---------------------------------------------------------分界線------------------------------------------------------------------------
有題目描述可知：
該題 典型的拓撲排序。
1、在 拓撲排序時，當出現同時可訪問兩個節點或多個節點時，則資訊不完全。
2、當出現有環時，則有衝突，判斷是否有環可以在拓撲排序完時判斷是否有點未訪問。因為若無環，每個點都可以訪問。
不多說了，附上我的AC程式碼：
1 #include <queue>
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 using namespace std;
6 const int MAXN=2500;
7 int n,m,cas,id;
8
9 struct edge
10 {
11int v,nx;
12 }set[MAXN];
13 int head[30],d[30],ok[30],write[30];
14 queue<int> Q;
15
16 void Addedge(int u,int v)
17 {
18id++;set[id].v=v;set[id].nx=head[u];
19head[u]=id;
20 }
21
22 int bfs()
23 {
24cas=0;
25while(!Q.empty())Q.pop();
26int out=1,t,u;t=0;
27for(int i=1;i<=n;i++)
28if(!ok[i])
29{
30t++;Q.push(i);
31}
32if(t>1)out=0;
33while(!Q.empty())
34{
35u=Q.front();Q.pop();t=0;write[++cas]=u;
36for(int k=head[u];k>0;k=set[k].nx)
37{
38ok[set[k].v]--;
39if(!ok[set[k].v])
40{
41t++;Q.push(set[k].v);
42}
43}
44if(t>1)out=0;
45}
46for(int i=1;i<=n;i++)if(ok[i]>0){out=-1;break;}
47return out;
48 }
49
50 char print()
51 {
52for(int i=1;i<=cas;i++)printf("%c",(char)(write[i]+64));
53return '.';
54 }
55
56 int main()
57 {
58while(~scanf("%d%d",&n,&m))
59{
60memset(write,0,sizeof(write));
61memset(d,0,sizeof(d));
62memset(head,-1,sizeof(head));
63id=0;
64int now=0,loca;
65char a,b,c;
66if(n==0 && m==0)break;
67for(int i=1;i<=m;i++)
68{
69cin>>a>>b>>c;
70if(b=='<')
71{
72Addedge(a-64,c-64);d[c-64]++;
73}
74else
75{
76Addedge(c-64,a-64);d[a-64]++;
77}
78if(now==0)
79{
80for(int j=1;j<=n;j++)ok[j]=d[j];
81now=bfs();loca=i;
82}
83}
84if(now==0)puts("Sorted sequence cannot be determined.");
85else if(now==1)printf("Sorted sequence determined after %d relations: ",loca),printf("%c\n",print());
86else printf("Inconsistency found after %d relations.\n",loca);
87}
88return 0;
89 }
記得點贊歐。