题目链接:Link
Problem
Solution
暴力枚举边数,用传递闭包判断当前状态即可。
Code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cassert>
using namespace std;
const int maxn=30;
int n,m,d[maxn][maxn],in[maxn];
char s[maxn][4];
int solve(int m)
{
memset(d,0,sizeof(d));
for(int i=1;i<=m;i++) d[s[i][0]-'A'+1][s[i][2]-'A'+1]=1;
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
d[i][j]|=d[i][k]&d[k][j];
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(d[i][j]&&d[j][i]) { printf("Inconsistency found after %d relations.
",m); return 1; }
for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(i!=j&&d[i][j]+d[j][i]==0) return 0;
printf("Sorted sequence determined after %d relations: ",m);
for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(d[i][j]) in[j]++;
queue<int> Q;
for(int i=1;i<=n;i++) if(in[i]==0) Q.push(i);
assert(Q.size()==1);
while(Q.size())
{
int u=Q.front(); Q.pop();
printf("%c",char(u-1+'A'));
for(int i=1;i<=n;i++) if(d[u][i])
{
d[u][i]=0;
in[i]--;
if(in[i]==0) Q.push(i);
}
}
puts(".");
return 1;
}
int main()
{
#ifdef local
freopen("pro.in","r",stdin);
#endif
while(scanf("%d%d",&n,&m)==2&&n+m)
{
for(int i=1;i<=m;i++) scanf("%s",s[i]);
for(int i=1;i<=m;i++) if(solve(i)) goto nxt;
puts("Sorted sequence cannot be determined.");
nxt:continue;
}
return 0;
}