Sorting It All Out

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K

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.
分析：优先判断有没有环；
　　　然后判断能不能全部排好序；
　　　最后才是条件不全；
　　　前两个利用拓扑排序，入度为0入队列，同一时间有>1个在队列中，条件不全；
　　　拓扑完还有入度的为环，不成立，输出；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e4+10;
const int N=5e4+10;
const int M=N*10*10;
using namespace std;
inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline void umax(ll &p,ll q){if(p<q)p=q;}
inline void umin(ll &p,ll q){if(p>q)p=q;}
int n,m,k,t,du[26],dudu[26],ret[27];
vi e[26];
char op[5];
int gao()
{
    int i,j,ok=1;
    ret[0]=0;
    queue<int>p;
    rep(i,0,n-1){du[i]=dudu[i];if(!du[i])p.push(i);}
    while(!p.empty())
    {
        if(p.size()>1)ok=0;
        int q=p.front();
        p.pop();
        ret[++ret[0]]=q;
        for(int i=0;i<e[q].size();i++)
        {
            int x=e[q][i];
            if(--du[x]==0)p.push(x);
        }
    }
    rep(i,0,n-1)if(du[i])return 2;
    return ok&&ret[0]==n;
}
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&k))
    {
        bool flag=false;
        if(!n&&!k)break;
        rep(i,0,n-1)dudu[i]=0,e[i].clear();
        rep(i,1,k)
        {
           scanf("%s",op);
           if(op[1]=='>')swap(op[0],op[2]);
           e[op[0]-'A'].pb(op[2]-'A');
           dudu[op[2]-'A']++;
           if(flag)continue;
           int x=gao();
           if(x==1)
           {
               printf("Sorted sequence determined after %d relations: ",i);
               rep(j,1,ret[0])printf("%c",ret[j]+'A');
               puts(".");
               flag=true;
           }
           else if(x==2)printf("Inconsistency found after %d relations.
",i),flag=true;
        }
        if(!flag)puts("Sorted sequence cannot be determined.");
    }
    return 0;
}