POJ 1094：Sorting It All Out（拓扑排序+传递闭包）

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19834 Accepted: 6784

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 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 0

Sample Output

Sorted sequence determined after 4 relations: ABCD.Inconsistency found after 2 relations.Sorted sequence cannot be determined.

Source

East Central North America 2001

对前n个字母，输入m组确定他们之间的大小关系，从而判断有没有唯一的序列。

第一种输出，在前t(t<=m)已能确定一个序列

第二种:不能确定唯一的序列

第三种：不存在拓扑排序

思路：拓扑排序+floyd。先用拓扑排序判断在当前情况下，有没有环的存在，如果有环的存在则证明m组关系是有矛盾的，

即可输出3。若没有环，那没接下来就是判断能否构成唯一的有序序列，用floyd传递闭包，如果能构成唯一的有序序列，

则顶点1~n，必分布着0~n-1个入度，才能保证0个入度的顶点排在第1位，1个入度的顶点排在第2位（因为只有1个顶点比它大），

以此类推。有一点要注意下，就是要每输入一次判断一下，如果有输出了，以后就只输入就好，不用第二次输出。

源代码：

#include<iostream>using namespace std;bool graph[30][30];int n,m;int path[30];bool TopSort(){int i,j,k;int indegree[30];for(i=0;i<n;i++){indegree[i]=0;for(j=0;j<n;j++)indegree[i] += graph[j][i];//cout<<i<<":"<<indegree[i]<<endl;}for(k=0;k<n;k++){for(i=0;i<n && indegree[i]!=0;i++);if(i==n) return false;indegree[i] = -1;path[k] = i;for(j=0;j<n;j++)indegree[j] -= graph[i][j];}return true;}bool Floyd(){int i,j,k;bool map[30][30];    for(i=0;i<n;i++)for(j=0;j<n;j++)map[i][j]=graph[i][j];/*cout<<"1:"<<endl;    for(i=0;i<n;i++){for(j=0;j<n;j++)cout<<graph[i][j]<<" ";cout<<endl;}*/for(k=0;k<n;k++)for(i=0;i<n;i++)for(j=0;j<n;j++)if(i!=k && k!=j && map[i][k] && map[k][j])map[i][j] = 1;/*cout<<"2:"<<endl;for(i=0;i<n;i++){for(j=0;j<n;j++)cout<<map[i][j]<<" ";cout<<endl;}*/bool vis[30],finish=false;memset(vis,false,sizeof(vis));int totalIndegree = 0;for(i=0;i<n;i++){totalIndegree = 0;for(j=0;j<n;j++)totalIndegree += map[j][i];//cout<<":"<<totalIndegree<<endl;if(vis[totalIndegree])return false;vis[totalIndegree] = true;}return true;}int main(){char c1,c2,c3;bool finish;while(scanf("%d%d",&n,&m)!=EOF){if(n == 0 && m == 0)break;finish = false;memset(graph,0,sizeof(graph));for(int i=1;i<=m;i++){cin>>c1>>c2>>c3;if(finish) continue;//if it finished before,just input the data and do nothingint u = c1-'A';int v = c3-'A';if(graph[u][v] == true) continue;   //this relationship have already existed!!!!graph[u][v] = true;if(TopSort() == 0)//TopSort firstly,if a loop exist,inconsistency found{finish = true;//set the flag finish as trueprintf("Inconsistency found after %d relations.\n",i);continue;}if(Floyd())//judge..{finish = true;printf("Sorted sequence determined after %d relations: ",i);for(int j=0;j<n;j++)printf("%c",'A'+path[j]);cout<<"."<<endl;continue;}}if(!finish)printf("Sorted sequence cannot be determined.\n");}}