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

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total 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 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.

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;
}