Sorting It All Out
| Time Limit: 1000MS | Memory Limit: 10000K |
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 0Sample Output
Sorted sequence determined after 4 relations: ABCD. Inconsistency found after 2 relations. Sorted sequence cannot be determined.
题意：给出m个比较式，问能否能确定有序数列，若不能，是条件不足还是因为有环存在。
解题思路: 拓扑排序，（一个字母看成一个点） 没输入一个式子进行一次拓扑排序，若同时有多个入度为0的点，说明条件不足，若没有入度为0的点后，确定顺序的点少于n个，说明有环存在。要先判断是否有环再判断是否能确定顺序，前者是后者的真子集
ACCode：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
int n,_in[27],tmp[27],ans[27];
vector< int> _out[27];
bool vis[27][27],flag;
void init()
{
for(int i=0;i<n;i++) {
_in[i] = 0;
_out[i].clear();
}
memset(vis,false,sizeof vis);
flag = false;
}
int Topo()
{
queue< int> q;
int len=0;
for(int i=0;i<n;i++) {
if(_in[i] == 0)
q.push(i);
}
bool uflag = false;
while(!q.empty()) {
if(q.size() > 1) uflag=true;; //还不能判断
int t = q.front();
q.pop();
ans[++len] = t;
for(int i=0;i<_out[t].size();i++) {
_in[_out[t][i]]--;
if(_in[_out[t][i]] == 0)
q.push(_out[t][i]);
}
}
// 要先判断是否有环再判断是否能确定顺序，前者是后者的真子集
if(len < n) return 1;
if(uflag) return 2;
return 3;
}
int main()
{
int m,t;
char ord[4];
while(~scanf("%d%d%*c",&n,&m) && (n || m)) {
init();
for(int i=1;i<=m;i++) {
scanf("%s",ord);
if(flag) continue;
if(ord[1] == '<' && !vis[ord[0]-'A'][ord[2]-'A']) {
vis[ord[0]-'A'][ord[2]-'A'] = true;
_in[ord[2]-'A']++;
_out[ord[0]-'A'].push_back(ord[2]-'A');
}
else if(ord[1] == '>' && !vis[ord[2]-'A'][ord[0]-'A']) {
vis[ord[2]-'A'][ord[0]-'A'] = true;
_in[ord[0]-'A']++;
_out[ord[2]-'A'].push_back(ord[0]-'A');
}
memcpy(tmp,_in,sizeof(_in));
int t = Topo();
memcpy(_in,tmp,sizeof(tmp));
if(t == 1) {
printf("Inconsistency found after %d relations.\n",i);
flag = true;
}
else if(t == 3) {
printf("Sorted sequence determined after %d relations: ",i);
for(int i=1;i<=n;i++)
printf("%c",char(ans[i]+'A'));
printf(".\n");
flag = true;
}
}
if(!flag)
printf("Sorted sequence cannot be determined.\n");
}
return 0;
}