Sort AcWing

Sort AcWing

Description

• Given n variables, m inequalities.

  Inequality between transitive, that is, if A> B and B> C, then A> C.

  It is determined whether inequality m contradictory.

  If there is contradiction, the minimum value of t is determined, just before t satisfies inequality can be determined contradiction between inequality.

  Without contradiction, it is determined whether the inequality m can determine the relationship between each of the variables.

  If, then obtain the minimum value of t, t satisfy the inequality just before can determine the magnitude relationship between the variables of each pair.

Input

• Test input comprising a plurality of sets of data.

  Each test, the first line contains two integers n and m.

  Subsequently m rows, each row containing an inequality, less than all of the inequality relationship.

  When the input line 00, represents the input termination.

Output

• A data output representing the results of each line.

  The result may be one of the following three:

  1, "Sorted sequence determined after t relations:. Yyy ... y", which refers to all yyy ... y variables in ascending order.
  2, "Sorted sequence can not be determined.".
  3, "Inconsistency found after t relations .".

Data Size

• 2 ≦ n- ≤ 26, the variable may only capital letters A ~ Z.

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.

answer:

• Floyd / topological sorting.
• This problem is obviously a transitive closure of the title, first i <j written as f (i, j) = 1. Flyod then passed under the closure. After two cycles pass enumeration i and j, if present, f (i, j) and F (j, i) are 1, a contradiction. When f (i, j) and f (j, i) is 0, indicating that can not be determined. The earliest position may be determined and then with diethyl points.
• Well but I tried to write a topology.
• First of all i <j, then even an edge of i to j. Clearly, then determine the relationship must be a chain. Then each added a set of relationships, do a topology. If there is a ring, a direct contradiction to return. In the internal topology function in each round if there is> 1 point appears in the queue, can not be determined explained. After the final finish topology, it can be determined and no explanation is contradictory, if it comes to a point and then all the points already, you can return.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#define N 105
using namespace std;

struct E {int next, to;} e[100005];
int n, m, num;
int h[N], in[N], t[N], seq[N];
bool vis[N], tag[N];
bool rel[N][N];
vector<int> a;

void add(int u, int v)
{
    e[++num].next = h[u];
    e[num].to = v;
    h[u] = num;
}

int topsort()
{
    queue<int> que;
    int flag = 0, cnt = 0;
    for(int i = 1; i <= 26; i++) t[i] = in[i];
    for(int i = 0; i < a.size(); i++)
        if(!in[a[i]]) que.push(a[i]);
    while(que.size())
    {
        if(que.size() > 1) flag = 1;
        int now = que.front(); que.pop();
        seq[++cnt] = now;
        for(int i = h[now]; i != 0; i = e[i].next)
        {
            in[e[i].to]--;
            if(!in[e[i].to]) que.push(e[i].to);
        }
    }
    for(int i = 1; i <= 26; i++) in[i] = t[i];
    if(cnt != a.size()) return -1;
    if(!flag && cnt == n) return 1;
    else return 0;
}

void ini()
{
    a.clear(), num = 0;
    memset(h, 0, sizeof(h));
    memset(vis, 0, sizeof(vis));
    memset(in, 0, sizeof(in));
    memset(rel, 0, sizeof(rel));
}

int main()
{
    while(scanf("%d%d", &n, &m) == 2)
    {
        if(!n && !m) break;
        ini();
        for(int i = 1; i <= m; i++)
        {
            char c[6]; scanf("%s", c);
            int u = c[0] - 'A' + 1, v = c[2] - 'A' + 1;
            if(!rel[u][v])
            {
                rel[u][v] = 1;
                if(!vis[u]) vis[u] = 1, a.push_back(u);
                if(!vis[v]) vis[v] = 1, a.push_back(v);
                add(u, v), in[v]++;
            }
            int ans = topsort();
            if(ans == -1)
            {
                printf("Inconsistency found after %d relations.\n", i);
                for(int j = i + 1; j <= m; j++) scanf("%s", c);
                break;
            }
            else if(!ans && i == m) printf("Sorted sequence cannot be determined.\n");
            else if(ans == 1)
            {
                printf("Sorted sequence determined after %d relations: ", i);
                for(int j = 1; j <= n; j++) printf("%c", (char)(seq[j] - 1 + 'A'));
                printf(".\n");
                for(int j = i + 1; j <= m; j++) scanf("%s", c);
                break;
            }
        }
    }
    return 0;
}