joj 1089 &&zoj 1060&&poj 1094 以及wa的分析和数据

这道题着实让无数人纠结，wa了n次n天，poj的discuss有大量的数据，wa的同学可以试一试。

结果我把数据都过了还是wa，后来在discuss的帮助下终于ac了，判断是否有环要最先判断，例如如下的数据：

4 4

A<B

C<B

D<B

B<A

总的思想是拓扑排序，我的方法比较笨，先不完全拓扑排序判断有无环，在完全拓扑排序判断是否满足了条件:

#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
int map[27][27];
int rudu[27];
int r_mark[27];
int fff;  //一个分量中如度为0的点的个数
int chu;  //一个分量中入度为零得点
int dfs(int m)
{
    int i;
   
    if(r_mark[m]==1)
        return 0;
    else
        r_mark[m]=1;
    if(rudu[m]==0)
    {
        fff++;
        chu=m;
    }
    int k=0;
    for(i=0;i<27;i++)
       if((map[i][m]||map[m][i])&&r_mark[i]==0)
          k+=dfs(i);
    return k+1;
}
int main()
{
    int m,n;
    //freopen("out","w",stdout);
    while(scanf("%d%d",&m,&n),m||n)
    {
        int i,j;
        memset(map,0,sizeof(map));
        memset(rudu,0,sizeof(rudu));
        int rrr=0;
        for(i=0;i<n;i++)
        {
            char str[10];
            scanf("%s",str);
            if(map[str[0]-'A'][str[2]-'A']==1)
                 continue;
            map[str[0]-'A'][str[2]-'A']=1;
            rudu[str[2]-'A']++;
            int sum=0,k;
            if(rrr)
               continue;
            for(j=0;j<m;j++)
            {
                if(rudu[j]==0)
                {
                   sum++;
                   k=j;
                }
            }
            if(sum==0)
            {
               printf("Inconsistency found after %d relations.\n",i+1);
               rrr=1;
               continue;
            }
            int ww_rudu[27];                  //不完全拓扑排序证明无环
                for(j=0;j<27;j++)
                    ww_rudu[j]=rudu[j];
            j=0;
            while(ww_rudu[j]!=0)
               j++;
            k=j;
            int count=0;
            while(1)
            {
                 count++;
                 ww_rudu[k]=-1;
                 for(j=0;j<m;j++)
                 {
                        if(map[k][j])
                        {
                             ww_rudu[j]--;
                        }
                 }
                 j=0;
                 while(ww_rudu[j]!=0&&j<m)
                     j++;
                 k=j;
                 if(j==m)
                     break;
            }
            if(count<m)
            {
               printf("Inconsistency found after %d relations.\n",i+1);
               rrr=1;
               continue;
            }
            int jj;
            memset(r_mark,0,sizeof(r_mark));
            for(jj=0;jj<m;jj++)
            {
                if(r_mark[jj]==1)
                    continue;
                fff=0;
                queue<int> q;
                int mm=dfs(jj);
                if(mm==1)
                    continue;
                if(fff==0)
                {
                    printf("Inconsistency found after %d relations.\n",i+1);
                    rrr=1;
                    break;
                }
                int cc=0;
                int mark=0;
                int w_rudu[27];
                for(j=0;j<27;j++)
                    w_rudu[j]=rudu[j];
                int k=chu;
                q.push(k);
                int wa=0;
                if(fff==1&&mm==m)     //完全拓扑排序求解最终问题
                while(1)
                {
                    cc++;
                    //cout<<"aaaaaa"<<endl;
                    int sum=0,kk;
                    for(j=0;j<m;j++)
                    {
                       if(map[k][j])
                       {
                           w_rudu[j]--;
                           if(w_rudu[j]==0)
                           {
                                sum++;
                                kk=j;
                           }
                       }
                    }
                    if(sum==1)
                    {
                        k=kk;
                        q.push(k);
                    }
                    else
                        break;
                }
                // cout<<mark<<" "<<cc<<" "<<mm<<" "<<fff<<endl;
                if(cc==m&&wa==0)
                {
                    printf("Sorted sequence determined after %d relations: ",i+1);
                    while(!q.empty())
                    {
                        printf("%c",q.front()+'A');
                        q.pop();
                    }
                    printf(".\n");
                    rrr=1;
                }
            }
        }
        if(rrr==0)
        {
            printf("Sorted sequence cannot be determined.\n");
        }
    }
}