Sorting It All Out
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total 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 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.
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 26
vector<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;
}