| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 29182 | Accepted: 10109 |
Description
Input
Output
Sample Input
4 6 A<B A<C B<C C<D B<D A<B 3 2 A<B B<A 26 1 A<Z 0 0
Sample Output
Sorted sequence determined after 4 relations: ABCD. Inconsistency found after 2 relations. Sorted sequence cannot be determined.
Source
#include<stdio.h>
const int N = 30;
int mapt[N][N],in[N],n,m;
int tope(int tm)
{
int path[N],k=0,l=0,uncertain=0;
for(int i=0;i<n;i++)
if(in[i]==0)
path[k++]=i,in[i]=-1;
while(l<k)
{
int s=path[l++];
if(k-l!=0)
uncertain=1;
for(int j=0;j<n;j++)
if(in[j]>0&&mapt[s][j])
{
in[j]--;
if(in[j]==0)
path[k++]=j,in[j]=-1;
}
}
int flag=0;
if(k!=n)//说明有环,矛盾
printf("Inconsistency found after %d relations.\n",tm),flag=1;
else if(uncertain==0)
{
printf("Sorted sequence determined after %d relations: ",tm);
for(int i=0;i<k;i++)
printf("%c",path[i]+'A');
printf(".\n");
flag=1;
}
else if(tm==m)//最后一次加入一个关系判断,可能的输出
printf("Sorted sequence cannot be determined.\n"),flag=1;
return flag;
}
int main()
{
int in_t[N];
char st[10];
while(scanf("%d%d",&n,&m)>0&&n+m!=0)
{
for(int i=0;i<n;i++)
{
in_t[i]=0;
for(int j=0;j<n;j++)
mapt[i][j]=0;
}
int flag=0;
for(int i=1;i<=m;i++)
{
scanf("%s",st);
if(flag)
continue;
int a=st[0]-'A';
int b=st[2]-'A';
if(mapt[a][b]==0)
{
mapt[a][b]=1; in_t[b]++;
for(int j=0;j<n;j++)
in[j]=in_t[j];
flag=tope(i);
}
}
}
}
POJ1094 Sorting It All Out(拓扑排序)每输入条关系判断一次
原文地址:http://blog.csdn.net/u010372095/article/details/45194521