POJ 1094 Sorting It All Out(トポロジーソート・判定+実装)
いくつかの異なる要素からなる昇順シーケンスは、すべての要素を小さい順にいくつかの小さい番号で並べ替えることができるシーケンスです.例えば、ソート後のシーケンスはA、B、C、Dであり、これはA各関係が小さいたびにトポロジーソートが行われ、衝突(すなわちループが発生する)または結果が決定されたときに後続の関係が処理される必要がないため、flagタグ結果が決定されたかどうかを使用することができます.
Sorting It All Out
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 "
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
#include
#include
#include
using namespace std;
const int N = 30;
int n, m, in[N];
char ans[N], q[N];
vector e[N];
int topoSort() //
{
int d[N], ret = 1;
memcpy(d, in, sizeof(d));
int front = 0, rear = 0, p = 0;
for(int i = 0; i < n; ++i)
if(d[i] == 0) q[rear++] = i;
while(front < rear)
{
if(rear - front > 1) ret = 0; //
int cur = q[front++];
ans[p++] = 'A' + cur;
for(int i = 0; i < e[cur].size(); ++i)
{
int j = e[cur][i];
if((--d[j]) == 0) q[rear++] = j;
}
}
if(p < n) return -1; //
ans[p] = 0;
return ret;
}
int main()
{
char a, b;
while(scanf("%d%d", &n, &m), n || m)
{
for(int i = 0; i < n; ++i) e[i].clear();
memset(in, 0, sizeof(in));
int flag = 0;
for(int i = 0; i < m; ++i)
{
scanf(" %c Sorting It All Out
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 "
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
Sample Output
Sorted sequence determined after 4 relations: ABCD.
Inconsistency found after 2 relations.
Sorted sequence cannot be determined.