poj 1094 Sorting It All Out （拓扑排序）

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27929 Accepted: 9655

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.

思路：根据矛盾优先原则，即只要出现矛盾就结束程序，输出答案；然后再没有矛盾的基础上进行拓扑排序看能否得出它们之间关系；最后，若以上两者均没有结果，则是条件不足。同时，我们要输入一个关系进行一次判断。

在程序中，每次拓扑后若满足合法的拓扑条件，就记录答案。最后判断是否有矛盾关系即可。

#include"stdio.h"#include"string.h"#include"vector"#include"queue"#include"iostream"#include"algorithm"using namespace std;#define N 1000#define M 26vector<int>g[M];     //记录拓扑关系int pre[M],n,m,flag,t;int mark[M];        标记字符是否出现及是否使用过char ans[M+5];      记录答案void top_sort(int k){    t=k+1;                 int i,j,u,v,tmp;    queue<int>q;    memset(pre,0,sizeof(pre));    memset(mark,0,sizeof(mark));    for(i=0;i<n;i++)    {        if(g[i].size()!=0)            mark[i]=1;        //等于1代表该字符出现        for(j=0;j<g[i].size();j++)        {            v=g[i][j];            mark[v]=1;            pre[v]++;        }    }    j=0;    while(1)    {        tmp=0;        for(i=0;i<n;i++)        {            if(pre[i]==0&&mark[i]==1)            {                mark[i]=-1;      //字符已经用过一次                u=i;                         q.push(u);                tmp++;            }        }        if(tmp==0)       //排序结束标志            break;        else if(tmp==1) //条件合法，记录答案            ans[j++]='A'+u-0;        while(!q.empty())        {            u=q.front();            q.pop();            for(i=0;i<g[u].size();i++)            {                v=g[u][i];                pre[v]--;            }        }    }    for(tmp=i=0;i<n;i++)        if(pre[i])            tmp++;    if(tmp>0)      //出现该情况肯定是出现矛盾         flag=2;    else if(j==n)        flag=1;    ans[j]='\0';}int main(){    char s[10];    int i,u,v;    while(scanf("%d%d",&n,&m),n||m)    {        flag=0;        for(i=0;i<M;i++)            g[i].clear();        for(i=0;i<m;i++)        {            scanf("%s",s);            if(!flag)            {                u=s[0]-'A';                v=s[2]-'A';                g[u].push_back(v);                top_sort(i);        //进行拓扑排序            }        }        if(flag==1)            printf("Sorted sequence determined after %d relations: %s.\n",t,ans);        else if(flag==2)            printf("Inconsistency found after %d relations.\n",t);        else            printf("Sorted sequence cannot be determined.\n");    }    return 0;}