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.
题目大意：
给你n个字母和m个约束条件，让你给出最后能否找到一个序列符合。
解题思路：
拓扑排序。
这道题目= =非常的坑，比如如果你在前面以及确定了一个唯一的序列能够符合约束，后面再给出一个矛盾的条件是无效的。。。
如果排除这些之后，这道题目还是非常简单的。直接对于每次约束进行拓扑排序就可以了。
代码：
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <functional>
using namespace std;
#define pb push_back
const int maxn = 30;
vector<int> ans, tot, vec[maxn];
int in[maxn], vis[maxn], tmp[maxn], cnt;
int toposort(int n){
priority_queue<int, vector<int>, greater<int> > q;
while(!q.empty()) q.pop();
for(int i = 0; i < maxn; ++i) tmp[i] = in[i];
int rec = 0;
for(int i = 0; i < 26; ++i)
if(!tmp[i] && vis[i]) {
q.push(i);
++rec;
}
if(rec == 1) rec = 0;
while(!q.empty()){
int p = q.top();
tot.pb(p); q.pop();
int v, len = vec[p].size();
for(int i = 0; i < len; ++i){
v = vec[p][i];
--tmp[v];
if(!tmp[v]) q.push(v), ++rec;
}
if(rec == 1) rec = 0;
}
if( !rec && tot.size() == n) return 1;
else if( tot.size() == cnt ) return 0;
else return -1;
}
int main()
{
// freopen("test.in", "r+", stdin);
// freopen("test.out", "w+", stdout);
char a, b;
int n, m, x, y, res, flag;
while(~scanf("%d%d", &n, &m)){
if(n == 0 && m == 0) break;
cnt = 0; flag = 0;
memset(in, 0, sizeof(in));
memset(vis, 0, sizeof(vis));
for(int i = 0; i < maxn; ++i) vec[i].clear();
for(int i = 1; i <= m; ++i) {
scanf(" %c<%c", &a, &b);
x = a - 'A';
y = b - 'A';
++in[y];
if(!vis[x]) ++cnt, vis[x] = 1;
if(!vis[y]) ++cnt, vis[y] = 1;
vec[x].pb(y);
if(flag == 0){
tot.clear();
int k = toposort(n);
if(k == 1) {
ans = tot;
flag = 1; res = i;
}else if(k == -1){
flag = -1; res = i;
}
}
}
if(flag == 1){
printf("Sorted sequence determined after %d relations: ", res);
for(int i = 0; i < ans.size(); ++i){
printf("%c", 'A'+ans[i]);
}
puts(".");
}else if(flag == 0){
printf("Sorted sequence cannot be determined.
");
}else{
printf("Inconsistency found after %d relations.
", res);
}
}
return 0;
}