POJ1094-Sorting It All Out

题目链接：点击打开链接

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38356		Accepted: 13525

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.

题目大意：给出以下关系，①处理到第几个时能确定所有的关系，②处理到第几个时能发现矛盾，③没有矛盾所有点都能确定

思路：先是想到并查集，但是谁比谁大不好比较。后来构建有向图，每读入一组关系，就拓扑排序一次。

AC代码：

#include<iostream>
#include<cstdio>
#include<set>
#include<queue>
#include<string.h>
using namespace std;

int n, m, cnt;

int indegree[1010], temp[1010];

vector<int> G[1010];//邻接表
queue<int> q;
char ans[31];

bool check(int x, int y) {
	for(int i = 0; i < G[x].size(); i++) {
		if(G[x][i] == y)
			return true;
	}
	return false;
}

int topo() {
	for(int i = 0; i < n; i++) {//这里是跑所有的点
		if(!indegree[i])
			q.push(i);
	}
	bool unsure = false;//不确定的情况
	cnt = 0;
	while(!q.empty()) {//因为是跑所有的点，一旦出现有的点没有涉及，入读为0的点必然大于1
		if(q.size() > 1)
			unsure = true;
		int u = q.front(); //取队列首个顶点u
		ans[cnt++] = u + 'A';
		q.pop();
		for(int i = 0; i < G[u].size(); i++) {
			int v = G[u][i];
			indegree[v]--; //入度减一 
			if(indegree[v] == 0) {//入度为0，入队 
				q.push(v);
			}
		} 
	}
	if(cnt < n)//有环，矛盾
		return 2;
	if(unsure)
		return 3;
	return 1;
}

char s[3];
int main() {
	int step, flag;
	while(~scanf("%d %d", &n, &m)) {
		memset(G, 0, sizeof(G));
		while(!q.empty()) q.pop();
		if(n == 0 && m == 0)
			break;
		bool ok = false;
		memset(indegree, 0, sizeof(indegree));
		for(int i = 0; i < m; i++) {
			scanf("%s", s);
			if(ok)//就算知道了结果，也要读入所有数据
				continue;
			int l = s[0]-'A';
			int r = s[2]-'A';
			if(!check(l, r)) {//有过了，就不用再加入了
				G[l].push_back(r);
				indegree[r]++;
			}
			memcpy(temp, indegree, sizeof(indegree));//把当前入度情况记录下来
			flag = topo();//拓扑判断一下
			memcpy(indegree, temp, sizeof(temp));//再把入读情况放进入，进行下一条的时候，上个入读情况还是保留的
			if(flag != 3) {
				step = i+1;//记录当前是第几条关系
				ok = true;
			}
		}//三种情况
		if(flag == 2) {
			printf("Inconsistency found after %d relations.
", step);
		} else if(flag == 3){
			printf("Sorted sequence cannot be determined.
");
		} else {
			ans[cnt] = '';
			printf("Sorted sequence determined after %d relations: %s.
", step, ans);
		} 
	}	
}