Sorting It All Out
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 26801 | Accepted: 9248 |
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.
Source
拓扑排序的应用。參考http://www.cnblogs.com/pushing-my-way/archive/2012/08/23/2652033.html做的。
本题须要注意的问题非常多，有点“坑”。以下是从上面博文中转的，（）里面的内容是我自己加的。
题意：给你一些大写字母间的偏序关系，然后让你推断是否能唯一确定它们之间的关系，或者所给关系是矛盾的，或者到最后也不能确定它们之间的关系。
分析：
用拓扑排序：
1.拓扑排序能够用栈来实现，每次入栈的是入度为0的节点（也能够用队列，或者不使用队列和栈,循环n次，找入度为0的点）。
1.拓扑排序的结果一般分为三种情况：1、能够推断（拓扑排序有唯一的结果） 2、有环出现了矛盾(出现了没有入度为0的节点） 3、条件不足，不能推断.
2.这道题不仅须要推断这三种情况，并且还要推断在处理第几个关系时出现前两种情况，对于本道题来说三种情况是有优先级的。前两种情况是平等的谁先出现先输出谁的对应结果，对于第三种情况是在前两种情况下都没有的前提下输出对应结果的.
网上对于这道题的错误提示（须要注意）：
1.本题顺序：
a.先判有没有环，有环就直接输出不能确定；
b.假设没有环，那么就看会不会有多种情况，假设有多种情况就再读下一行；假设所有行读完还是有多种情况，就是确定不了；
c.假设最后没有环，也不存在多种情况（即每次取出来入度为零的点仅仅有一个），那么才是答案；
2.有答案就先出答案，无论后面的会不会矛盾什么的；
3.假设在没有读全然部输入就能出答案，一定要把剩下的行都读完。
代码：
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stack>
#include <queue>
using namespace std;
int indegree[30];//保存入度
int graph[30][30];//是否有边
char output[30];//输出可确定序列
bool ok;//能够被确定
bool dilemma;//有环，矛盾
bool no;//不能被确定
char c1,c,c2;//输入
int topo(int n)
{
int in[30];
for(int i=0;i<n;i++)
in[i]=indegree[i];//使用备用数组进行拓扑排序
stack<int>s;//入度为0的点进栈
for(int i=0;i<n;i++)
if(!in[i])
s.push(i);
bool flag=0;//栈里面入度为0的元素大于一个时，不确定的拓扑排序
int cnt=0;//入度为0的元素个数，也是输出序列里面的个数
while(!s.empty())
{
if((s.size())>1)
flag=1;//不确定
int first=s.top();
s.pop();
output[cnt++]=first+'A';//放入输出序列里面
for(int i=0;i<n;i++)
if(graph[first][i])//与入度为0的元素相连的元素
{
in[i]--;
if(in[i]==0)
s.push(i);//入栈
}
}
if(cnt!=n)//假设没有环的话，序列里面的元素个数肯定等于输入的元素个数，就算在某个元素未输入之前，它的入度也初始化为0
return 2;//有环
else if(flag==1)//不确定的拓扑排序
return -1;
return 1;
}
int main()
{
int n,m;
while(cin>>n>>m&&(n||m))
{
ok=0;dilemma=0;no=0;
memset(indegree,0,sizeof(indegree));
memset(graph,0,sizeof(graph));
for(int i=1;i<=m;i++)
{
cin>>c1>>c>>c2;//当出现矛盾或者通过一些条件可被确定序列，剩下的输入条件就不须要再处理了
if(!ok&&!dilemma)//没有环，没有确定的拓扑排序，这里的拓扑排序必须输入的字母都有。比方输入ABCD 那么仅仅有AB不是确定的
{
int t1=c1-'A';
int t2=c2-'A';
if(graph[t2][t1])//双向边，有环，出现矛盾
{
cout<<"Inconsistency found after "<<i<<" relations."<<endl;
dilemma=1;//出现矛盾
continue;
}
if(!graph[t1][t2])
{
graph[t1][t2]=1;
indegree[t2]++;//入度++
}
int ans=topo(n);//确定返回1，有环返回2
if(ans==2)
{
cout<<"Inconsistency found after "<<i<<" relations."<<endl;
dilemma=1;
continue;
}
if(ans==1)
{
cout<<"Sorted sequence determined after "<<i<<" relations: ";
for(int k=0;k<n;k++)
cout<<output[k];
cout<<"."<<endl;
ok=1;
}
}
}
if(!ok&&!dilemma)
cout<<"Sorted sequence cannot be determined."<<endl;
}
return 0;
}