POJ1094:Sorting It All Out(拓扑排序)

本文介绍了一种使用拓扑排序算法来确定一组字母之间的大小顺序的方法。针对输入的一系列字母及其相互间的大小关系,该算法能够有效地判断是否可以确定这些字母的确切顺序,并在可能的情况下输出该顺序。

Sorting It All Out
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 35361 Accepted: 12417

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.

Source


题意:给N个字母,M个大小关系,问能否确定这N个字母的大小顺序,输出有三种:通过前X个关系能确定,通过前X个关系发现矛盾,所有关系都不能确定他们的大小顺序。

思路:这题怎么没给M范围,每增加一个关系都进行一次拓扑排序即可,处理起来不算难。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <vector>
# include <algorithm>
# define pb push_back
using namespace std;
const int maxn = 30;
vector<int>v[maxn];
int in[maxn], In[maxn], q[maxn<<1], vis[maxn], ans[maxn];
int n, m, flag, flag2, cur, id[28];
bool cmp(int x, int y){return ans[x]<ans[y];}
void check()
{
    int l=0, r=0, icount=0;
    memset(ans, 0, sizeof(ans));
    for(int i=0; i<n; ++i)  In[i] = in[i];
    for(int i=0; i<n; ++i) if(vis[i] && In[i]==0) q[r++] = i, ans[i]=1, ++icount;
    while(l<r)
    {
        int u = q[l++];
        for(int j=0; j<v[u].size(); ++j)
        {
            if(ans[v[u][j]]) {flag = 1; return;}
            if(--In[v[u][j]] == 0)//只要它入度不为0,说明还有比它等级小的,先不放进队列。
            {
                ++icount;
                ans[v[u][j]] = ans[u]+1;
                q[r++] = v[u][j];
            }
        }
    }
    if(icount < cur) flag = 1;//icount记录拓扑排序遍历过的节点数,小于当前已经出现的节点数,说明存在有向环。
    if(cur >= n)
    {
        int tmp[28]={0};
        for(int i=0; i<n; ++i)
            if(++tmp[ans[i]] > 1) return;//判断这N个节点的“等级”有无重复。
        flag2 = 1;
    }
}
int main()
{
    //freopen("in.txt","w",stdout);
    while(~scanf("%d%d",&n,&m),n+m)
    {
        char c, d;

        flag = flag2 = cur = 0;
        memset(vis, 0, sizeof(vis));
        memset(in, 0, sizeof(in));
        for(int i=0; i<n; ++i) id[i]=i, v[i].clear();
        for(int i=1; i<=m; ++i)
        {
            getchar();
            scanf("%c<%c",&c,&d);
            if(flag || flag2) continue;
            if(++vis[c-'A'] == 1) ++cur;
            if(++vis[d-'A'] == 1) ++cur;
            v[c-'A'].pb(d-'A');
            ++in[d-'A'];
            check();
            if(flag) {printf("Inconsistency found after %d relations.\n",i);}
            else if(flag2)
            {
                printf("Sorted sequence determined after %d relations: ",i);
                sort(id, id+n, cmp);
                for(int i=0; i<n; ++i) printf("%c",id[i]+'A');
                printf(".\n");
            }
        }
        if(!flag&&!flag2) printf("Sorted sequence cannot be determined.\n");
    }
    return 0;
}


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