[乱搞] BZOJ 4436 [Cerc2015]Kernel Knights

来源：互联网发布：北京谷歌seo推广编辑：程序博客网时间：2024/05/20 06:23

大概类似拓扑排序找入度为0的点然后删
最后剩下偶环任选一边就好了

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline void write(int x){  if (x>=10) write(x/10);  putchar('0'+x%10);}const int N=200005;int n,a[N];int tag[N],deg[N];int Q[N],l,r;int main(){  int x;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); n<<=1;  for (int i=1;i<=n;i++)    read(a[i]),deg[a[i]]++;  for (int i=1;i<=n;i++) if (!deg[i]) Q[++r]=i;  while (l<r){    int u=Q[++l]; tag[u]=1;    if (tag[a[u]]==0){      tag[a[u]]=2;      if (!(--deg[a[a[u]]]))    Q[++r]=a[a[u]];    }  }  for (int i=1;i<=n/2;i++) if (!tag[i]) tag[i]=1;  for (int i=1;i<=n;i++) if (tag[i]==1) write(i),putchar(' ');  return 0;}

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