POJ 1094 Sorting It All Out(拓扑排序变形)

最新推荐文章于 2019-09-03 09:49:38 发布

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本文介绍了一种复杂的拓扑排序问题，通过反向建图的方法解决排序确定性问题。文章详细阐述了实现步骤及注意事项，包括如何检测环路、确定排序完成的条件以及处理无法完全排序的情况。

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Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 30293		Accepted: 10496

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.


可以说这是一个很阴险的拓扑排序，我是反向建图做的，有三点需要注意：
1.如果有环的存在，要输出当前语句的下标（关键是这个）
2.如果已经排好序，则要在输入完成后输出顺序，并输出排序时所用到的语句的最小下标
3.如果元素不够，不足以排序，就另外输出

所以输出元素数组要一直更新！

ac代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stack>
#include<iostream>
#include<algorithm>
#define MAXN 50
#define INF 0xfffffff
#define max(a,b) a>b?a:b
#define min(a,b) a>b?b:a
using namespace std;
int pri[MAXN][MAXN];
int dis[MAXN];
int s[MAXN];
int n;
int topo()
{
	int i,j,k,num,temp=2; 
	int cas[MAXN];
	int ans=0;
	for(i=1;i<=n;i++)
	cas[i]=dis[i];
	for(i=0;i<n;i++)
	{
		num=0;
		for(j=1;j<=n;j++)
		{
			if(cas[j]==0)
			{
				k=j;
				num++;//记录当前最优位置的元素
			}
		}
		if(num==0)//如果成环
		return 0;
		if(num>1)//如果最优位置的元素不止一个，表明不足以排序，则标记返回，继续输入
		temp=1;
		s[ans++]=k;
		cas[k]=-1;
		for(j=1;j<=n;j++)
		{
			if(pri[k][j])
			{
				cas[j]--;
			}
		}
	}
	return temp;
}
int main()
{
	int i,j,m;
	char a[5];
	int k,bz,flag;
	while(scanf("%d%d",&n,&m)!=EOF,n||m)
	{
		bz=0;
		flag=0;
		memset(dis,0,sizeof(dis));
		memset(pri,0,sizeof(pri));
		for(i=1;i<=m;i++)
		{
			scanf("%s",a);
			if(bz||flag)//如果已经排好序或者成环了，就不用执行后面的了
			continue;
			pri[a[0]-'A'+1][a[2]-'A'+1]=1;//反向建图
			dis[a[2]-'A'+1]++;
		    int q=topo();
		    if(q==0) 
		    {
		    	printf("Inconsistency found after %d relations.\n",i);
		    	bz=1;
		    }
		    if(q==2)
			{
				flag=i;//记录最后排好序的语句下标
			}
		}
		if(flag)
		{
			printf("Sorted sequence determined after %d relations: ",flag);
			for(i=0;i<n;i++)
			printf(i==n-1?"%c.\n":"%c",s[i]+'A'-1);
			bz=1;
		}
		if(bz==0)
		printf("Sorted sequence cannot be determined.\n");
	}
	return 0;
}