Sorting It All Out
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 38379 | Accepted: 13534 |
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.
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.
Source
思路：就是利用拓扑排序，当条件不断增加时，依次判断，看是否能排好，同时要注意是否有环，注意在条件不足不能判断出先后顺序的情况下，依然要判断是否有环；
其实有一点有疑问，就是在前面条件充足，可以排好序之后即没有对后续的进行判断了，但是如果后面的条件使得前面的成环呢？貌似样例没有这个，可以直接ac
#include <iostream>
#include <cstring>
using namespace std;
int in[30]; //入度
int out[30]; //输出序列
int map[30][30]; //有向图
int TopoSort(int n){
int ct = 0, cpin[30];
int flag = 1;
for(int i = 0; i < n; i++){
cpin[i] = in[i];
}
while(ct < n){
int p = -1, m = 0;
for(int i = 0; i < n; i++){
if(cpin[i] == 0){
p = i;
m++;
}
}
if(m == 0){
return -1; //有环
}
if(m > 1){
// return 0; //无法判断
flag = 0; //得出不能判断只有还有检测，因为可能后面可能有环
}
out[ct++] = p;
cpin[p] = -1;
for(int i = 0; i < n; i++){
if(map[p][i]){
cpin[i]--;
}
}
}
return flag; //排好序了
}
int main(){
std::ios::sync_with_stdio(false);
int n, m;
char x, y;
char ch;
while(cin >> n >> m && (n != 0 && m != 0)){
memset(in, 0, sizeof(in));
memset(map, 0, sizeof(map));
int flag = 1;
for(int i = 0; i < m; i++){
cin >> x >> ch >> y;
if(flag == 0)
continue; //注意就算前面已经判断出结果，也要输入完，但不需要判断了
if(map[x - 'A'][y - 'A'] == 0){
map[x - 'A'][y - 'A'] = 1;
in[y - 'A']++;
int s = TopoSort(n);
if(s == -1){
cout << "Inconsistency found after ";
cout << i + 1;
cout << " relations." << endl;
flag = 0;
}
if(s == 1){
cout << "Sorted sequence determined after ";
cout << i + 1;
cout << " relations: ";
for(int i = 0; i < n; i++){
cout << char(out[i] + 'A');
}
cout << "." << endl;
flag = 0;
}
}
}
if(flag){
cout << "Sorted sequence cannot be determined." << endl;
}
}
return 0;
}