Sorting It All Out
时间限制：3000 ms | 内存限制：65535 KB
难度：3
- 描述
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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.
- 输入
- 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.
- 输出
- 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. - 样例输入
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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
- 样例输出
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Sorted sequence determined after 4 relations: ABCD. Inconsistency found after 2 relations. Sorted sequence cannot be determined.
题目大意：给你n个点，给你m条边代表大小关系。问你在第几条边加入后有矛盾（有环）或能确定关系，或者不能确定关系。
解题思路：首先每次加入一条边，就用floyd传递闭包，之后再判断是否形成环。如果没有环，就判断是否能确定唯一大小关系，这里有一个重要的判断条件即如果所有的结点的度等于n-1，则拓扑排序记录路径。#include<bits/stdc++.h> using namespace std; int Map[50][50],indegree[50],outdegree[50]; char S_ord[50]; bool floyd(int n){ for(int k=0;k<n;k++){ //传递闭包 for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ if(Map[i][k]&&Map[k][j]) Map[i][j]=1; } } } for(int i=0;i<n;i++) //判断是否形成环 if(Map[i][i]) return 1; return 0; } bool calcu_is_ord(int n){ //计算目前是否有序 memset(indegree,0,sizeof(indegree)); memset(outdegree,0,sizeof(outdegree)); for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ if(Map[i][j]){ indegree[j]++; outdegree[i]++; } } } for(int i=0;i<n;i++){ if(indegree[i]+outdegree[i]!=n-1){ /*如果所有结点都满足入度加出度等于结点总数减一，说明已经有序。因为如果有序，必然 会有入度为0~n-1，相应的出度为n-1~0。所以只要所有的结点度都为n-1，则说明已经有序。 */ return 0; } } return 1; } void topo_sort(int n){ //拓扑排序求大小顺序 int que_[50],vis[50],top=0,cnt=0,u; for(int i=0;i<n;i++){ if(indegree[i]==0){ que_[++top]=i; } } memset(vis,0,sizeof(vis)); while(top){ u=que_[top--]; vis[u]=1; S_ord[cnt++]=u+'A'; for(int i=0;i<n;i++){ if(!vis[i]&&Map[u][i]){ indegree[i]--; } if(!vis[i]&&indegree[i]==0){ que_[++top]=i; } } } S_ord[cnt++]=' '; } int main(){ int n,m; char str[10]; while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){ memset(Map,0,sizeof(Map)); int flag_cir=0,flag_ord=0; //记录在第几组关系输入时形成环或有序 for(int i=1;i<=m;i++){ scanf("%s",str); Map[str[0]-'A'][str[2]-'A']=1; if(flag_cir||flag_ord) continue; if(floyd(n)){ flag_cir=i;continue;} else if(calcu_is_ord(n)){topo_sort(n);flag_ord=i;continue;} } if(flag_cir) printf("Inconsistency found after %d relations. ",flag_cir); else if(flag_ord){ printf("Sorted sequence determined after %d relations: %s. ",flag_ord,S_ord); }else{ printf("Sorted sequence cannot be determined. "); } } return 0; }