POJ 1094 Sorting It All Out 拓扑排序+Floyd算法

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 24697 Accepted: 8557

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.

这道题虽然是一道拓扑排序的基本题，但是依然浪费了主页君一天来攻击这道题，这道题其实意思很简单，更具上面给出的关系求出一个整体关系，即给出N个字母，再给出字母之间大小关系，然后如果可以从大到小排序，则输出整体排序，如果对应关系矛盾，输出有矛盾，如果有不止一个对应关系，则输出存在多组对应关系，这道题其实运用拓扑算法，并且需要对拓扑算法进行一定修改，并且运用floyd看题目中是否有矛盾，判定方法就是求出全源对应关系，之后如果相同节点有大小对应关系，则证明题目存在矛盾，之后对数列带入拓扑算法，判断当寻找父节点为0的根节点时，发现有多个根节点，则证明对应关系不止一种，直接跳出函数继续更新路径后重新带拓扑排序，如果没有再发现根节点但是已排序节点数小于总结点数，则证明存在环路，即有矛盾，输出存在矛盾，之后的拓扑模板则不需要继续修改了，直接得到拓扑序列，并且输出即可AC，这道题其实也是道难度相对较大的题，各种关系量要注意分析的步骤，否则很容易因为粗心造成一定麻烦。

下面是AC代码：

#include<cstdio>#include<iostream>#include<cstring>using namespace std;int G[30][30],ans[30],degree[30],into[30],visit[30];int Floyd_Washall(int n){int i,j,k;    for (k=0;k<n;k++)        for (i=0;i<n;i++)            for (j=0;j<n;j++)                if (G[i][j]==0&&G[i][k]==1&&G[k][j]==1)                      G[i][j]=1;for(i=1;i<=n;i++)if(G[i][i]==1)return 1;return 0;}int Toplogical_sort(int n){        int i,j,top,count;     bool p=true;     top=0; memset(visit,0,sizeof(visit)); memset(degree,0,sizeof(degree)); memset(ans,0,sizeof(ans)); for(i=0;i<n;i++) for(j=0;j<n;j++) if(G[i][j]==1) degree[j]++; for(i=0;i<n;i++) { count=0; for(j=0;j<n;j++) if(degree[j]==0&&visit[j]==0) { count++; top=j; } if(count>=2) return 0; else if(count==0) return 1; ans[i]=top; visit[top]=1; for(j=0;j<n;j++) degree[j]--; } return 2;} int main(){int n,m,i,a,b,flag;char str[400][5];while(1){flag=0;memset(G,0,sizeof(G));cin>>n>>m;if(n==0&&m==0)break;for(i=0;i<m;i++)scanf("%s",str[i]);for(i=0;i<m;i++){a=str[i][0]-'A';b=str[i][2]-'A';G[a][b]=1;flag=Floyd_Washall(n);if(flag==1)break;flag=Toplogical_sort(n);if(flag==2)break;}if(flag==1)printf("Inconsistency found after %d relations.\n",i+1);else if(flag==0)printf("Sorted sequence cannot be determined.\n");else{printf("Sorted sequence determined after %d relations: ",i+1);for(i=0;i<n;i++)printf("%c",ans[i]+'A');printf(".\n");}}return 0;}