poj 1094 Sorting It All Out（拓扑排序）

poj 1094 Sorting It All Out（拓扑排序）2014-10-25 shuangde800 链接：

http://poj.org/problem？id=1094

题目：

Sorting It All OutTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21532 Accepted: 7403

DescriptionAn 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.

InputInput 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.

OutputFor 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.

分析：

这题典型的拓扑排序，但是有点变化。

题目样例的三种输出分别是：

1. 在第x个关系中可以唯一的确定排序，并输出。

2. 在第x个关系中发现了有回环（Inconsisitency矛盾）

3.全部关系都没有发现上面两种情况，输出第3种.

那么对于给定的m个关系，一个个的读进去，每读进去一次就要进行一次拓扑排序，如果发现情况1和情况2，那么就不用再考虑后面的那些关系了，但是还要继续读完后面的关系（但不处理）。如果读完了所有关系，还没有出现情况1和情况2，那么就输出情况3.

本栏目更多精彩内容：http://www.bianceng.cn/Programming/sjjg/

拓扑排序有两种方法，一种是算法导论上的，一种是用贪心的思想，这题用贪心的思想做更好。

贪心的做法：

1. 找到所有入度为0的点，加入队列Q

2.取出队列Q的一个点，把以这个点为起点，所有它的终点的入度都减1. 如果这个过程中发现经过减1后入度变为0的，把这个点加入队列Q。

3.重复步骤2，直到Q为空。

这个过程中，如果同时有多个点的入度为0，说明不能唯一确定关系。

如果结束之后，所得到的经过排序的点少于点的总数，那么说明有回环。

题目还需要注意的一点：如果边（u,v）之前已经输入过了，那么之后这条边都不再加入。

代码：

#include<cstdio>#include<cstring>#include<vector>#include<queue>using namespace std;const int N = 105;int n,m,in[N],temp[N],Sort[N],t,pos, num;char X, O, Y;vector<int>G[N];queue<int>q;void init（）{memset（in, 0, sizeof（in））;for（int i=0; i<=n; ++i）{G[i].clear（）;}}inline bool find（int u,int v）{for（int i=0; i<G[u].size（）; ++i）if（G[u][i]==v）return true;return false;}int topoSort（）{while（！q.empty（））q.pop（）;for（int i=0; i<n; ++i）if（in[i]==0）{q.push（i）;}pos=0;bool unSure=false;while（！q.empty（））{if（q.size（）>1） unSure=true;int t=q.front（）;q.pop（）;Sort[pos++]=t;for（int i=0; i<G[t].size（）; ++i）{if（--in[G[t][i]]==0）q.push（G[t][i]）;}}if（pos<n） return 1;if（unSure）return 2;return 3;}int main（）{int x,y,i,flag,ok,stop;while（~scanf（"%d%d%*c",&n,&m））{if（！n||！m）break;init（）;flag=2;ok=false;for（i=1; i<=m; ++i）{scanf（"%c%c%c%*c", &X,&O,&Y）;if（ok） continue; // 如果已经判断了有回环或者可唯一排序，不处理但是要继续读x=X-"A", y=Y-"A";if（O=="<"&&！find（y,x））{G[y].push_back（x）;++in[x];}else if（O==">"&&！find（x,y））{G[x].push_back（y）;++in[y];}// 拷贝一个副本，等下用来还原in数组memcpy（temp, in, sizeof（in））; flag=topoSort（）;memcpy（in, temp, sizeof（temp））;if（flag！=2）{stop=i;ok=true;}}if（flag==3）{printf（"Sorted sequence determined after %d relations: ", stop）;for（int i=pos-1; i>=0; --i）printf（"%c",Sort[i]+"A"）;printf（".
"）;}else if（flag==1）{printf（"Inconsistency found after %d relations.
",stop）;}else{printf（"Sorted sequence cannot be determined.
"）;}}return 0;}