poj 1094

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这题数据量较小，用了比较暴力的方法

A<B看作A到B的有向路径，每一条路径再分别连接两点前后的路径

当路径数满足 n*（n-1）/2是得出拓扑结构

#include<iostream>  #include <string>   #include<vector>  #include<algorithm>  #include<set>  #include<fstream>#include<cmath>  using namespace std;  #define lch(i) ((i)<<1)  #define rch(i) ((i)<<1|1)  #define sqr(i) ((i)*(i))  #define pii pair<int,int>  #define mp make_pair  #define FOR(i,b,e) for(int i=b;i<=e;i++)  #define ms(a)   memset(a,0,sizeof(a))  const int maxnum = 30;  const int inf = 100005;int road[maxnum][maxnum];bool vis[maxnum],vis1[maxnum];int seq[maxnum];int n,m;int main()  {  //ifstream fin("G:/1.txt");char str[4];int u,v,flag1,flag3,sum;while(1){scanf("%d%d",&n,&m);//fin>>n>>m;if(n==0&&m==0)break;flag1=flag3=sum=0;ms(road);ms(vis1);ms(seq);FOR(i,1,n)road[i][i]=1;FOR(i,1,m){//fin>>str;scanf("%s",str);if(flag1||flag3)continue;u=str[0]-'A'+1;v=str[2]-'A'+1;if(road[v][u]) {flag3=i;continue;}if(road[u][v]) continue;FOR(j,1,n){FOR(k,1,n){if(road[j][u]&&road[v][k]&&!road[j][k]){road[j][0]++;sum++;road[j][k]=1;}}}if(sum==n*(n-1)/2)flag1=i;}if(flag1==0&&flag3==0){printf("Sorted sequence cannot be determined.\n");}else if(flag3){printf("Inconsistency found after %d relations.\n",flag3);}else if(flag1){printf("Sorted sequence determined after %d relations: ",flag1);FOR(i,1,n){seq[road[i][0]]=i;}for(int i=n-1;i>=0;i--){printf("%c",seq[i]-1+'A');}printf(".\n");}else printf("!\n");}return 0;}

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