POJ - 1094 Sorting It All Out

题目意思比较简单易懂，给他t个关系式（A<B的形式），确定n个连续的字母是不是有序，无法确定还是矛盾

分析：比较容易想到拓扑排序。比如A<B可以看成，A的入度为0，B的入度为1，然后这样的式子有t个，也就是说，当某一个为字母的入度为0的时候这数就要入队列，并记下这个数（目前最小的数），然后跟比他的大的入度都要-1，但是确定时哪一种情况需要小心。有序的情况比较好确定、独立，矛盾跟无法确定的情况需要注意下，是无法的确定的情况只有在矛盾的情况是不成立的

附上自己的里程悲

#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define f first
#define s second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>PLL;
typedef pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const double inf=1e8+100;
const double eps=1e-8;
const ll maxn =1e3+200;
const int N = 1e4+10;
const ll mod=1000007;
//
int in[maxn],num[maxn],G[maxn][maxn];
int n,t;
//
int toposort() {
	queue<int>q;
	int tmp[maxn];
	memcpy(tmp,in,sizeof(in));
	bool flag=0;
	int cnt=0;
	for(int i=0; i<n; i++) {
		if(!tmp[i]) {
			q.push(i);
		}
	}
	while(!q.empty()) {
		if(q.size()>1) {
			flag=1;
		}
		int tp=q.front();
		q.pop();
		num[cnt++]=tp;
		for(int i=0; i<n; i++) {
			if(G[tp][i]==1&&!(--tmp[i])) {
				q.push(i);
			}
		}
	}
	if(cnt!=n)
		return 2;
	if(flag)
		return 1;
	return -1;
}
void solve() {
	int i,j,tt=1;
	while(cin>>n>>t) {
		if(n==0&&t==0)
			break;
		memset(G,0,sizeof(G));
		memset(in,0,sizeof(in));
		bool circle=0,order=0;
		int flag,k;
		for(i=0; i<t; i++) {
		//	cout<<(int)circle<<" "<<(int)order<<" "<<flag<<endl;
			char a,b,eq;
			cin>>a>>eq>>b;
			if(circle||order)
				continue;
			if(G[b-'A'][a-'A']) {
				circle=1;
				//	cout<<"circle"<<endl;
				cout<<"Inconsistency found after "<<i+1<<" relations."<<endl;
				continue;
			}
			if(!G[a-'A'][b-'A']) {
				G[a-'A'][b-'A']=1;
				in[b-'A']++;
			}
			flag=toposort();
			if(flag==2) {
				circle=1;
				//	cout<<"circle"<<endl;
				cout<<"Inconsistency found after "<<i+1<<" relations."<<endl;
				continue;
			} else if(flag==-1) {
				order=1;
				k=i;
			}
		}
		if(order) {
			//	cout<<"order"<<endl;
			cout<<"Sorted sequence determined after "<<k+1<<" relations: ";
			for(i=0; i<n; i++) {
				cout<<(char)(num[i]+'A');
			}
			cout<<"."<<endl;
		} else if(!circle&&flag==1) {
			//	cout<<"no det"<<endl;
			cout<<"Sorted sequence cannot be determined."<<endl;
		}
	}
}


int main() {
	ios_base::sync_with_stdio(false);
#ifdef debug
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	cin.tie(0);
	cout.tie(0);
	solve();
	/*
		#ifdef debug
			fclose(stdin);
			fclose(stdout);
			system("out.txt");
		#endif
	*/
	return 0;
}