拓扑排序---(poj 1094)

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal 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 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 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;}