poj1094 拓扑排序

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 24092 Accepted: 8321

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.

拓扑排序！详见代码#include<iostream>#include<cstring>using namespace std;int n,m;  //n结点下限，m关系对char top_out[26];  //排序输出列表int po=0;  //输出列表的指针typedef class degree{public:int in;      //入度char to[26];   //记录指向的所有顶点，以便删除出度的操作int pt;    //数组to的指针};int top_sort(degree alph[],bool mark[],int num){/*假设图G的当前子图为F*/memset(top_out,'\0',sizeof(top_out));po=0;int del_n=0;int zero=0;  //记录图F中入度为0的结点个数for(int i='A';i<'A'+n;i++)if(mark[i] && !alph[i].in)zero++;bool flag=false;while(zero>0){if(zero>1)  //图F的无前驱结点的个数不唯一，排序无法确定flag=true;  //考虑到"矛盾"的优先性，避免在多个0入度结点情况下，最后一步输入刚好出现环（此时为矛盾）                //所以这里先不返回值，而是先标记，执行拓扑，根据情况决定返回值for(int k='A';k<='A'+n;k++)   //寻找图F的唯一的前驱结点if(mark[k] && !alph[k].in){mark[k]=false;       //删除图F的唯一无前驱结点kdel_n++;            //记录删除的结点数top_out[po++]=k;    //k记录到排序输出列表for(int i=0;i<alph[k].pt;i++)   //删除结点k的所有出度边alph[ alph[k].to[i] ].in--;break;}zero=0;for(int j='A';j<='A'+n;j++)if(mark[j] && !alph[j].in)zero++;}    if(flag && del_n==num)return 3;if(del_n<num)   //说明图F存在有向环,矛盾，与0入度结点的多少无关。因为矛盾优先return 2;if(!flag && del_n==num && del_n<n)  //图F能排序，但不能确定图G是否能排序，还需继续输入观察return 3;if(!flag && del_n==n)    //图G能排序return 1;}int main(void){int num;      //标记前n个字母出现个数,用于最终检查是否前n个字母均已被读入     //*_t[]是用于备份的额外数组bool mark['Z'+1],mark_t['Z'+1];  //标记当前图G所使用的字母（结点）degree alph['Z'+1],alph_t['Z'+1];   while(true){/*Input*/cin>>n>>m;if(!n||!m)break;/*Initial*/memset(mark,false,sizeof(mark));memset(mark_t,false,sizeof(mark_t));num=0;for(int k='A';k<'A'+n;k++){alph[k].in=alph_t[k].in=0;alph[k].pt=alph_t[k].pt=0;memset(alph[k].to,'\0',sizeof(alph[k].to));memset(alph_t[k].to,'\0',sizeof(alph_t[k].to));}/*Structure Maps*/char x,symbol,y;  //临时变量bool flag=false;bool sign=false;int value;   //记录拓扑返回的值int step;   //记录当前情况发生的步骤for(int pair=1;pair<=m;pair++){cin>>x>>symbol>>y;if(x>='A'+n || y>='A'+n)  //当输入的结点不在前n个字母范围内时sign=true;       //不再进行拓扑,单纯检查后续输入是否把前n个字母都输入了                 //为了区分非前n个字母的字母的输入时间，是在确认了排序或矛盾之前还是之后                     //在确认 排序或矛盾之前:flag=false,sign=true                     //在确认 排序或矛盾之后:flag=true,sign=trueif(!mark[x] && x<'A'+n)num++;if(!mark[y] && y<'A'+n)num++;if(!flag && !sign){value=0;    mark[x]=mark[y]=true;        //顶点标记    mark_t[x]=mark_t[y]=true;    alph[y].in++;                //入度标记    alph_t[y].in++;    alph[x].to[ alph[x].pt++ ]=y;        //指向标记 & 指针移动    alph_t[x].to[ alph_t[x].pt++ ]=y;/*Top-Sort & Sign*/value=top_sort(alph_t,mark_t,num);  //每次输入后图都被更新，要重新拓扑    if(value==1)       //排序确认{     step=pair;    //记录确认排序的位置    flag=true;    //不再对后续输入处理}    else if(value==2)  //矛盾{    step=pair;    //记录矛盾发生的位置    flag=true;    //不再对后续输入处理}    else if(value==3 && pair<m)  //排序（暂时）无法确认，需继续处理后续输入     for(int k='A';k<'A'+n;k++)        //数据还原{    mark_t[k]=mark[k];    alph_t[k].in=alph[k].in;}if(pair==m && value==0)value=3;}if(sign && !flag && num==n)  //在确认 排序或矛盾之前,当存在有非前n个字母的结点时的"矛盾"{step=pair;value=2;}else if(sign && !flag && pair==m && num<n) //在确认 排序或矛盾之前,当存在有非前n个字母的结点时的"无法确认排序"value=3;}if(value==1){cout<<"Sorted sequence determined after "<<step<<" relations: ";for(int i=0;i<po;i++)cout<<top_out[i];cout<<'.'<<endl;}else if(value==2)cout<<"Inconsistency found after "<<step<<" relations."<<endl;else if(value==3)cout<<"Sorted sequence cannot be determined."<<endl;}return 0;}