（poj1094）Sorting It All Out

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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.

Hint

#include<cstdio>  
#include<cstring>  
#include<queue>  
#include<vector>  
using namespace std;  
const int maxn=26+5;  
vector<int> G[maxn];  
int in[maxn];  
int n,m;  
char order[1000][10];  
char ans[maxn];  
int cnt;
int topo ()
{int i;
cnt=0;bool p=true;
queue<int> q;  
    int in1[maxn];
memcpy(in1,in,sizeof(in));  
for(i=0;i<n;i++)
if(in1[i]==0)q.push(i);
while(!q.empty())
{
if(q.size()>1)p=false;
int u=q.front();q.pop();
ans[cnt++]=u+'A';
for(i=0;i<G[u].size();i++)
{int v=G[u][i];
if(--in1[v]==0)q.push(v);
}
}
int p1=0;
if(cnt<n) p1 = -1;
    else if(p==1) p1=1; 
    return p1;


}

int main()
{
while(scanf("%d%d",&n,&m)==2&&n)  
    {  int i;
        memset(in,0,sizeof(in));  
        for(i=0;i<n;i++) G[i].clear();  
        for(i=0;i<m;i++)  
            scanf("%s",order[i]);  
        int flag=0;  
        for(i=0;i<m;i++)  
        {  
            int u=order[i][0]-'A', v=order[i][2]-'A';  
            G[u].push_back(v);  
            in[v]++;  
            if( (flag=topo())!=0) break;  
        }  
        ans[n]=0;      
        if(flag==1) printf("Sorted sequence determined after %d relations: %s.\n",i+1,ans);  
        else if(flag==0) printf("Sorted sequence cannot be determined.\n");  
        else if(flag==-1) printf("Inconsistency found after %d relations.\n",i+1);  
    }  
    return 0;  
}  


第一步  ans  AB->CD;
第二步           ABC->D
第三步           ABCD

转载于:https://www.cnblogs.com/lengxia/p/4387802.html