POJ_1094_sorting it all out_拓扑排序

今天舍友回来，等会回去啊了他怎么样。

题意：
对于前n个大写字母，给出他们之间m个小于关系，要求确定他们是否存在矛盾，或者存在唯一的拓扑序，或者无法确定排序，当已输入的关系可以得出三种结论中的一种时，就得出结论，即使和后面的矛盾。

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.

这题一看就知道是拓扑排序，但他还要确定是否是唯一的排序，然后我个逗比就觉得不能用拓扑标准过程来了，先自己写了个拓扑排序，中间夹杂了很多操作，一直TLE，做了组大数据把操作数输出来一看，10^10以上哦凑。。。剪了剪枝还是10^9，然后就按拓扑标程写，又为了确定排序是否唯一加了一些东西，结果WA了。。想明白了怎么回事以后，突然想到，只要跑一个拓扑排序，再按照得出的拓扑序看一下是否每两个相邻的元素都存在边就行了。。。。。。。。。。

所以说，no zuo no die 还是自己太弱。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<string>using namespace std;#define mxn 30int n,m;bool mapp[mxn][mxn];int suc,fail;int flag[mxn],cnt;char ans[mxn];void init(){ans[n]=0;suc=fail=-1;memset(mapp,false,sizeof(mapp));}bool dfs(int now){flag[now]=-1;for(int i=0;i<n;++i)if(mapp[now][i])if(flag[i]==-1||flag[i]==0&&!dfs(i))return false;ans[--cnt]=now+'A';flag[now]=1;return true;}bool toposort(){memset(flag,0,sizeof(flag));cnt=n;for(int i=0;i<n;++i)if(!flag[i])if(!dfs(i))return false;return true;}bool check(){for(int i=0;i<n-1;++i)if(!mapp[ans[i]-'A'][ans[i+1]-'A'])return false;return true;}int main(){while(scanf("%d%d",&n,&m)!=EOF&&n&&m){init();for(int i=0;i<m;++i){char tem[5];int u,v;scanf("%s",tem);u=tem[0]-'A';v=tem[2]-'A';if(!mapp[u][v]&&suc==-1&&fail==-1){mapp[u][v]=true;if(!toposort())fail=i+1;elseif(check())suc=i+1;}}if(fail!=-1)printf("Inconsistency found after %d relations.\n",fail);else if(suc!=-1)printf("Sorted sequence determined after %d relations: %s.\n",suc,ans);elseputs("Sorted sequence cannot be determined.");}return 0;}