POJ-1094-Sorting It All Out（拓扑排序）

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本文介绍了ACM竞赛中POJ-1094题目，重点探讨了如何进行拓扑排序。关键在于判断图中是否存在环以及节点的顺序状态，一旦发现环或确定有序或无序，即可停止处理并输出结果。

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Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32037		Accepted: 11131

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.

Source

East Central North America 2001

注意：判断 1，构成环 2，有序 3，无序。

1,2成立立刻无视后续数据输出。

//#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int Scan()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')flag=1;
    else if(ch>='0'&&ch<='9')res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';
    return flag?-res:res;
}
void Out(int a){if(a>9)Out(a/10);putchar(a%10+'0');}
int n,m;
char str[5];
int map_[30][30],in[30],p[30],stack_[30];
int TopoSort()
{
    memcpy(p,in,sizeof(in));
    int top=0,flag=1;
    for(int i=1; i<=n; ++i)
    {
        int temp=0,poi;
        for(int j=1; j<=n; ++j)
        if(p[j]==0){temp++; poi=j;}
        if(!temp)return 0;//构成环
        if(temp>1)flag=-1;//无序
        stack_[top++]=poi;
        p[poi]=-1;
        for(int j=1; j<=n; ++j)
            if(map_[poi][j])p[j]--;
    }
    return flag;//有序
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(!n||!m)break;
        memset(in,0,sizeof(in));
        memset(map_,0,sizeof(map_));
        int flag=0;
        for(int i=1; i<=m; ++i)
        {
            scanf("%s",str);
            if(flag)continue;
            int a=str[0]-'A'+1;
            int b=str[2]-'A'+1;
            map_[a][b]=1;
            ++in[b];
            int poi=TopoSort();
            if(poi==0)
            {
                printf("Inconsistency found after %d relations.\n",i);
                flag=1;
            }
            else if(poi==1)
            {
                printf("Sorted sequence determined after %d relations: ",i);
                for(int j=0;j<n;j++)
                    printf("%c",stack_[j]+'A'-1);
                printf(".\n");
                flag=1;
            }
        }
        if(!flag)printf("Sorted sequence cannot be determined.\n");
    }
    return 0;
}