POJ1094 Sorting It All Out —— 拓扑排序
题目链接:http://poj.org/problem?id=1094
Sorting It All Out
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 35468 | Accepted: 12458 |
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.
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.
题解:
1.由于要输出当关系确定时是第几步,所以每一步都需要进行拓扑排序。(假如读完所有数据再进行拓扑排序,则无法确定关系确定时是第几步)。
2.在进行拓扑排序时(flag为状态类型,1表示冲突; 2表示不能确定; 3表示关系确定。 flag初始化为3):
(1) 如果入度为0的点的个数为0,则成环,可直接得出结果:冲突。直接return 1。
(2) 如果入度为0的点的个数为1,继续。
(3) 如果入度为0的点的个数大于1,则表明最小的那个不能确定。这时把flag改为2,然后继续。(为什么还要继续,不是可以直接return 2,表明关系还不能确定了吗? 虽然如此,但是继续拓扑排序下去,可能会发现环,即冲突。)
代码如下:
#include<stdio.h>//poj 1094 #include<string.h> #include<stdlib.h> int deg[27], sub_deg[27],edge[27][27],n,m; char a,b,c,ans[27]; int TopoSort() { int flag = 3; memcpy(sub_deg,deg,sizeof(deg)); for(int i = 0; i<n; i++) { int k, cnt = 0; for(int j = 0; j<n; j++) if(sub_deg[j]==0) cnt++, k = j; if(cnt==0) return 1; //成环 if(cnt>1) flag = 2; //最小的不能确定。但不能直接退出,因为接下来可能会出现环。 sub_deg[k] = -1; ans[i] = k+65; for(int j = 0; j<n; j++) if(edge[k][j]) sub_deg[j]--; } return flag; } int main() { int flag,step; while(scanf("%d%d",&n,&m) && (n||m)) { int finished = 0, flag, step; memset(deg,0,sizeof(deg)); memset(edge,0,sizeof(edge)); for(int i = 1; i<=m; i++) { getchar(); scanf("%c%c%c",&a,&c,&b); if(finished) continue; deg[b-65]++; edge[a-65][b-65] = 1; ans[n] = 0; flag = TopoSort(); if(flag==1 || flag==3) { step = i; finished = 1; } } if(flag==1) printf("Inconsistency found after %d relations. ",step); else if(flag==2) printf("Sorted sequence cannot be determined. "); else printf("Sorted sequence determined after %d relations: %s. ",step,ans); } return 0; }