[杂题] Codeforces 799E Round #413 E. Aquarium decoration

枚举下选了几个ab都喜欢的 没什么难度 就是细节贼多

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;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;}const int N=200005;int n,m,K,w[N],idx[N],back[N];ll cnt[N],sum[N];inline void add(int x,int r){  for (int i=x;i<=n;i+=i&-i)    cnt[i]++,sum[i]+=r;}inline ll query(ll *c,int x){  ll ret=0;  for (int i=x;i;i-=i&-i)    ret+=c[i];  return ret;}inline ll Query(int k){  int L=0,R=n+1,MID;  while (L+1<R)    if (query(cnt,MID=(L+R)>>1)<=k)      L=MID;    else      R=MID;  return query(sum,L);}bool cmp(int x,int y){  return w[x]<w[y];}int sx[N],icnt;int a[N],b[N],c[N],d[N];ll suma[N],sumb[N],sumc[N];int la[N],lb[N];ll ans=1LL<<60;int main(){  int x,k;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(m); read(K);  if (K>m) return printf("-1\n"),0;  for (int i=1;i<=n;i++) read(w[i]),idx[i]=i;  sort(idx+1,idx+n+1,cmp);  for (int i=1;i<=n;i++) back[idx[i]]=i;  read(x); if (x<K) return printf("-1\n"),0; while (x--) read(k),la[k]=1;  read(x); if (x<K) return printf("-1\n"),0; while (x--) read(k),lb[k]=1;  for (int i=1;i<=n;i++)    if (la[i] && lb[i]) c[++*c]=i;    else if (la[i]) a[++*a]=i;    else if (lb[i]) b[++*b]=i;    else d[++*d]=i;  if (*c+(K-*c)*2>m) return printf("-1\n"),0;  for (int i=1;i<=*d;i++) add(back[d[i]],w[d[i]]);  sort(a+1,a+*a+1,cmp); for (int i=1;i<=*a;i++) suma[i]=suma[i-1]+w[a[i]];  sort(b+1,b+*b+1,cmp); for (int i=1;i<=*b;i++) sumb[i]=sumb[i-1]+w[b[i]];  sort(c+1,c+*c+1,cmp); for (int i=1;i<=*c;i++) sumc[i]=sumc[i-1]+w[c[i]];  int C=max(m-*a-*b-*d,max(max(K-*a,K-*b),2*K-m));  C=max(C,0);  int pnt1=*a,pnt2=*b;  while (pnt1>0 && pnt1>=K-C+1) add(back[a[pnt1]],w[a[pnt1]]),pnt1--;  while (pnt2>0 && pnt2>=K-C+1) add(back[b[pnt2]],w[b[pnt2]]),pnt2--;  ll ret=sumc[C]+suma[max(K-C,0)]+sumb[max(K-C,0)]+Query(m-(C+max(K-C,0)*2));  ans=min(ans,ret);  for (C++;C<=min(m,*c);C++){    while (pnt1>0 && pnt1>=K-C+1) add(back[a[pnt1]],w[a[pnt1]]),pnt1--;    while (pnt2>0 && pnt2>=K-C+1) add(back[b[pnt2]],w[b[pnt2]]),pnt2--;    ret=sumc[C]+suma[max(K-C,0)]+sumb[max(K-C,0)]+Query(m-(C+max(K-C,0)*2));    ans=min(ans,ret);  }  printf("%I64d\n",ans);  return 0;}