poj1094 Sorting It All Out

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 36044 Accepted: 12680

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 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 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：给很多的两两数字之间的大小关系（e.g A<B ），然后问你能不能给出他们的总的排序。

题解：拓扑排序。每次输入一个关系，就进行一次拓扑排序：

1.如果排序唯一，那么输出第一种情况
2.如果出现了环，那么输出第二种情况
3.如果到了输入完毕后，排序仍不唯一，那么输出第三种情况 

一开始我让它中途出现了排序不唯一直接输出cannot be determined，后来发现sb了，直到所有东西都输完了才能确定能否唯一确定。情况1和2是一旦中途出现，就可以输出了。

代码如下：

#include<stdio.h>#include<queue>#include<vector>#include<cstring>#define SORTED 1#define INCON 2#define CANTDERTER 3using namespace std;const int MAX = 300;vector <int> G[MAX];int indegree[MAX];int temp[MAX]; //备份indegree int pos[MAX]; //输出 int n, m;int cnt;/*输入一个，就进行一次拓扑排序1.如果排序唯一，那么输出第一种情况2.如果出现了环，那么输出第二种情况3.如果到了输入完毕后，排序仍不唯一，那么输出第三种情况 */ int topo(){queue<int> q;cnt = 0;int unsure = 0;for(int i = 1;i <= n; ++i){if(!indegree[i]){q.push(i);}}while(!q.empty()){if(q.size() > 1)unsure = 1;int v = q.front();q.pop();pos[cnt++] = v;for(int i = 0;i < G[v].size();++i){if(!--indegree[G[v][i]]){q.push(G[v][i]);}}}if(cnt < n) return INCON;if(unsure) return CANTDERTER;return SORTED;} int main(){char ch1,ch2,ch3;int x,y;int ok, res;while(scanf("%d%d",&n,&m) == 2 && (n||m)){memset(pos,0,sizeof(pos));memset(indegree,0,sizeof(indegree));for(int i = 0;i < MAX; ++i)G[i].clear();ok = 0;for(int i = 1;i <= m; ++i){while(getchar()!='\n')continue;scanf("%c%c%c",&ch1,&ch3,&ch2);if(ok) continue;int x = ch1 - 'A' + 1, y = ch2 - 'A' + 1;G[x].push_back(y);++indegree[y];//备份indegreememcpy(temp,indegree,sizeof(indegree)); res = topo();//还原indegreememcpy(indegree,temp,sizeof(indegree)); if(res == SORTED) {ok = 1;printf("Sorted sequence determined after %d relations: ",i);for(int j = 0;j < cnt; ++j)printf("%c",pos[j]+'A'-1);printf(".\n"); }else if(res == INCON){ok = 1;printf("Inconsistency found after %d relations.\n",i);}else continue;}if(res == CANTDERTER)printf("Sorted sequence cannot be determined.\n");}}