POJ-1094 Sorting It All Out

Sorting It All Out

Time Limit: 1000MS

Memory Limit: 10000K

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.

————————————————————集训6.2的分割线————————————————————

前言：整整做了一天。拓扑排序本身并没有那么难，但是姿势不对，就没有可行性了。因为这个并不是输入所有的边之后再让你排序的模板题，而是每次输入一条边，都要重新进行拓扑排序，并且要判断这一次是否产生了唯一解或者是否成环。

题意：很久才弄清楚。输入数字N，（要把A～Z的前N个字母进行拓扑排序）。然后输入M，（存在M条边）。

如果在某一次输入之后产生环，输出是哪一次。

如果确定了完整且唯一的拓扑序，输出在第几次得到，以及这个序列。

如果既没有产生环，而且不能确定唯一的拓扑序，输出cannot。（神题啊）

思路：每一次输入之后，进行一次拓扑排序。正确姿势如下：

For(i = 0 ~ n) {//对每个顶点
	For(k = 0 ~ n) {
		找到一个入度为0且未曾访问的点cur;
	}
	For(与该点相连所有的边) {
		入度--;
	}
	vis[cur] = true;
}

这道神题的关键点是做判断的优先级。一定要先判断是否成环，然而想要判断是否成环则要等到在剩下的点中找不到入度为0的点的时候。而且初次判断结束后，不必再重新判断（Fu k）。

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <iostream>
using namespace std;
/****************************************/
const int N = 30, M = N*N;
bool vis[N], confuse, go_on;
int n, m, indeg[N], degree[N];
vector <int> G[N], ans;

void init()
{
	ans.clear();
	memcpy(degree, indeg, sizeof(indeg));
	memset(vis, 0, sizeof(vis));
	confuse = false;
}

int Topsort()
{
	init();
	for(int k = 0; k < n; k++) {
		int zero = 0, alph = 0, cur;
		for(int i = 0; i < n; i++) {
			if(vis[i]) continue;
			if(!degree[i]) {
				cur = i;
				zero++;
			}
		}
		if(zero) {
			for(int i = 0; i < G[cur].size(); i++) {
				degree[G[cur][i]]--;
			}
			vis[cur] = true;
			ans.push_back(cur);
		}
		if(zero == 0) 
			return 1;//最高优先级别，但是必须在所有0入度点访问完之后才能肯定
		if(zero > 1)
			confuse = true;//多个入度为0，继续访问入度为0的点
	}
	if(confuse) return 2;
	return 0;
}

void PR(int ret, int rela)
{
	if(ret == 1) {
		go_on = false;
		printf("Inconsistency found after %d relations.\n", rela);
	}
	else if(ret == 0) {
		go_on = false;
		printf("Sorted sequence determined after %d relations: ", rela);
		for(int j = 0; j < n-1; j++)
			printf("%c", ans[j]+'A');
		printf("%c.\n", ans[n-1]+'A');
	}
	else if(rela == m && ret == 2) {
		puts("Sorted sequence cannot be determined.");
	}
}

int main()
{
#ifdef J_Sure
	freopen("1094.in", "r", stdin);
	//freopen(".out", "w", stdout);
#endif
	while(scanf("%d%d", &n, &m),n) {
		for(int i = 0; i < N; i++) 
			G[i].clear();
		memset(indeg, 0, sizeof(indeg));
		go_on = true;
		char a, b; int u, v;
		int ret;
		for(int i = 1; i <= m; i++) {
			scanf(" %c<%c", &a, &b);
			u = a - 'A'; v = b - 'A';
			if(go_on) {
				indeg[v]++;
				G[u].push_back(v);
				ret = Topsort();
				PR(ret, i);
			}
		}
	}
	return 0;
}