POJ1094——Sorting It All Out

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 32845 Accepted: 11409

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    //0 0 时结束

Sample Output

Sorted sequence determined after 4 relations: ABCD.   //在第几个关系是就可以确定关系  输出关系Inconsistency found after 2 relations.       //在第几个关系时  出现矛盾Sorted sequence cannot be determined.       //无法确定关系

Source

East Central North America 2001

文字信息来自：http://www.cnblogs.com/pushing-my-way/archive/2012/08/23/2652033.html

题意：给你一些大写字母间的偏序关系，然后让你判断能否唯一确定它们之间的关系，或者所给关系是矛盾的，或者到最后也不能确定它们之间的关系。

分析：

用拓扑排序：

1.拓扑排序可以用栈来实现，每次入栈的是入度为0的节点。

1.拓扑排序的结果一般分为三种情况：1、可以判断 2、有环出现了矛盾 3、条件不足，不能判断.

2.这道题不仅需要判断这三种情况，而且还要判断在处理第几个关系时出现前两种情况，对于本道题来说三种情况是有优先级的。前两种情况是平等的谁先出现先输出谁的相应结果，对于第三种情况是在前两种情况下都没有的前提下输出相应结果的.

#include <iostream>#include <cstdlib>#include <stack>#include <cstdio>#include <cstring>using namespace std;const int N = 30;int n, m;int graph[N][N], indu[N], list[N];   //graph[i][j] ==1  表示  i<j   indu[i]表示 i 的入度， list储存输出顺序 int toopsort(){    int tmp[N];      //为了不影响indu数组   这里把indu数组复制到tmp数组中    memcpy(tmp,indu,sizeof(indu));    stack<int> s;    while(!s.empty())        s.pop();    for( int i = 0;i < n;i++ )    {        if( tmp[i]==0 )            s.push(i);            //如果入度为0的点大于1  则 不确定关系    }    int flag = 0;    int j = 0;    while(!s.empty())    {        if( s.size()>1 )         //判断<span style="font-family: Arial, Helvetica, sans-serif;">入度为0的点是否大于1</span>            flag = 1;        int t = s.top();        s.pop();        list[j++] = t;        for( int i = 0;i < n;i++ )        {            if( graph[t][i]==1 )     //t和i存在关系                if( --tmp[i]==0 )     // i 的入度减1,  此时  如果入度为0  则进栈                    s.push(i);        }    }    if( j!=n )     //you huan    //只有j==n的时候才说明不存在环        return 1;    else if( flag )  //no sure   //在没有环的情况下  判断时候能够确定大小关系        return 2;    else        return 0;      //上述两种情况都不存在}int main(){    while(scanf("%d%d",&n,&m)&&n&&m)    {        getchar();   //读掉回车键        char a, b;        int flag1 = 0;    //flag1  flag2   表示变量   控制输出；        int flag2 = 0;        memset(graph,0,sizeof(graph));     //初始化        memset(indu,0,sizeof(indu));        for( int i = 1;i <= m;i++ )        {            scanf("%c<%c",&a,&b);            getchar();      //读掉回车键            if( !flag1&&!flag2 )     //还没有输出过   进行下面的判断            {                if( graph[b-'A'][a-'A']==1 )       //存在环                {                    printf("Inconsistency found after %d relations.",i);                    flag1 = 1;     //表示变量flag1改变                    continue;                }                if( graph[a - 'A'][b - 'A']==0 )       //防止重复计算入度                {                    graph[a - 'A'][b - 'A'] = 1;                    indu[b-'A']++;                }                int tmp = toopsort();      //拓扑排序                if( tmp==1 )    //有环存在                {                    printf("Inconsistency found after %d relations.",i);                    flag1 = 1;                }                else if( tmp==0 )    //能够确定关系                {                    flag2=  1;     //标志变量flag2改变                    printf("Sorted sequence determined after %d relations: ",i);                    for( int j = 0;j < n;j++ )                    {                        printf("%c",list[j]+'A');                    }                    printf(".");                }            }        }        if( !flag1&&!flag2 )     //没有环  没有确定关系  就剩下无法确定了        {            printf("Sorted sequence cannot be determined.");        }        printf("\n");    }    return 0;}