ons esc def input miss with con nts following

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 39602		Accepted: 13944

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 分析： 有向圖判環：拓撲排序 無向圖判環：並查集 本題需要進行m次的拓撲排序 拓撲排序判環：拓撲之後，如果存在沒有入隊的點，那麽該點一定是環上的點 拓撲路徑的唯一性確定：隊中某一時刻存在兩個元素，則至少有兩條不同的拓撲路徑

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};

int getval()
{
    int ret(0);
    char c;
    while((c=getchar())==‘ ‘||c==‘\n‘||c==‘\r‘);
    ret=c-‘0‘;
    while((c=getchar())!=‘ ‘&&c!=‘\n‘&&c!=‘\r‘)
        ret=ret*10+c-‘0‘;
    return ret;
}

#define max_v 55
int indgree[max_v];
int temp[max_v];
int G[max_v][max_v];
int tp[max_v];
int n,m;
queue<int> q;
int tpsort()
{
    while(!q.empty())
        q.pop();
    for(int i=1;i<=n;i++)
    {
        indgree[i]=temp[i];
        if(indgree[i]==0)
            q.push(i);
    }

    int c=0,p;
    int flag=0;
    while(!q.empty())
    {
        if(q.size()>1)
            flag=1;
        p=q.front();
        q.pop();
        tp[++c]=p;
        for(int i=1;i<=n;i++)
        {
            if(G[p][i])
            {
                indgree[i]--;
                if(indgree[i]==0)
                    q.push(i);
            }
        }
    }
    /*
    拓撲完之後，存在沒有入隊的點，那麽該點就一定是環上的
    */
    if(c!=n)//存在環 
        return 1;
    else if(flag)//能拓撲但存在多條路
        return 0;
    return -1;//能拓撲且存在唯一拓撲路徑
}
int main()
{
    int x,y;
    char c1,c2;
    while(~scanf("%d %d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        memset(G,0,sizeof(G));
        memset(temp,0,sizeof(temp));
        memset(tp,0,sizeof(tp));
        int flag1=0,index1=0;//是否有環及環的位置
        int flag2=0,index2=0;//能否拓撲和拓撲的位置
        for(int i=1;i<=m;i++)
        {
            getchar();
            scanf("%c<%c",&c1,&c2);
            x=c1-‘A‘+1;
            y=c2-‘A‘+1;
            if(flag1==0&&flag2==0)
            {
                if(G[y][x])//環的一種情況
                {
                    flag1=1;
                    index1=i;
                    continue;
                }
                if(G[x][y]==0)//預防重邊
                {
                    G[x][y]=1;
                    temp[y]++;
                }
                int k=tpsort();
                if(k==1)//存在環
                {
                    flag1=1;
                    index1=i;
                    continue;
                }else if(k==-1)//存在唯一拓撲路徑
                {
                    flag2=1;
                    index2=i;
                }
            }
        }
        if(flag1==0&&flag2==0)
        {
            printf("Sorted sequence cannot be determined.\n");
        }else if(flag1)
        {
            printf("Inconsistency found after %d relations.\n",index1);
        }else if(flag2)
        {
            printf("Sorted sequence determined after %d relations: ",index2);
            for(int i=1;i<=n;i++)
            {
                printf("%c",tp[i]+‘A‘-1);
            }
            printf(".\n");//！！！註意還有個點...
        }
    }
    return 0;
}

POJ 1094 Sorting It All Out（拓撲排序+判環+拓撲路徑唯一性確定）