POJ1094：Sorting It All Out（拓扑排序）

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

East Central North America 2001

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;
}