拓扑排序---(poj 1094)

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36338		Accepted: 12786

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.

提示：一个基于拓扑排序的算法，就是输出输出和输入不好弄；

#include <iostream>
#include <cstring>
#include <cstdio>
#include <stack>
using namespace std;
const int N = 30;
int in[N]; //入度
bool Map[N][N]; //临接矩阵此题可以随便用
char out[N]; //拓扑排序输出数组

int topsort( int n ) {//传入参数为节点数；


    stack<int> q;
    int zeroflag = 0;
    int num = 0,temp[N];
    memcpy(temp,in,sizeof(in));//拷贝，为了将原来in里面的数据保存，目的就是这是为了下一次的拓扑能继续；
    for( int i = 0; i < n; i++ )
        if( !in[i] )
            q.push(i);

    while( !q.empty() ) {
        if( q.size() > 1 )//这就说明是无序的；然而题目要求是有序的；
            zeroflag = 1;//要注意在while里面判断 因为此过程中也可能出现入度0不止一个的情况；
        int k = q.top();
        q.pop();
        num++;
        out[num] = k + 'A'; //num计数 以存到out里面
        for(int i = 0; i < n; i++ )
            if(Map[k][i]==true && --temp[i] == 0)
                q.push(i);
    }


    if( num != n ) return 1;//表示有环
    if( zeroflag == 1 ) return 0;//多个点入度为零
    return -1;//表示无环
}

int main(){
    int n,m;

    char s[6];
    while(~scanf("%d%d", &n, &m) &&n&&m){


        int step = 0; //记录操作数
        int circleflag=0,orderflag=0;
        memset(Map,0,sizeof(Map));
        memset(in,0,sizeof(in));


        for( int i = 1; i <= m; i++ ){
            scanf("%s", s);
            if( circleflag == 0 && orderflag == 0 ){ //已经判出了 就不读了

                if(Map[s[2]-'A'][s[0]-'A'] == true){//有关系就为true;
                    circleflag=1; //有环了
                    printf("Inconsistency found after %d relations.\n", i);
                    continue ;
                }

                if(Map[s[0]-'A'][s[2]-'A'] == false){//读入第一组数据；
                    Map[s[0]-'A'][s[2]-'A'] = true;
                    in[s[2]-'A']++;//入读；
                }
                int k = topsort(n);//拓扑排序；

                if( k == 1 ){//有环；
                    circleflag=1;
                    printf("Inconsistency found after %d relations.\n", i);
                }
                else if( k == -1 ){//如果已经构成一个有向无环图；
                    orderflag=1;
                    step=i; //记录位置
                }
            }
        }

        if(circleflag == 0 && orderflag == 0)
            printf("Sorted sequence cannot be determined.\n");
        else if(orderflag == 1){
            printf("Sorted sequence determined after %d relations: ", step);
            for(int i=1; i<=n; i++)
                printf("%c", out[i]);
            printf(".\n");
        }
    }
    return 0;
}