题解
很容易求出在没有字典序最大的限制条件下的最多胜利场数。
这样就可以对于每一位放最优的解,怎么做,二分答案。
分两种情况,一种是当前一位是输的,一种是赢的,复杂度 \(\mathcal O(\rm nlog^2n)\) 卡卡常即可。
Code
#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
namespace IO{
char buf[1<<21],*p1=buf,*p2=buf;
#define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
struct nanfeng_stream{
template<typename T>inline nanfeng_stream &operator>>(T &x) {
ri f=1;x=0;register char ch=gc();
while(!isdigit(ch)) {if (ch=='-') f=0;ch=gc();}
while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
return x=f?x:-x,*this;
}
}cin;
}
using IO::cin;
namespace nanfeng{
#define FI FILE *IN
#define FO FILE *OUT
template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
static const int N=1e5+7;
int an[N],bn[N],tmx[N],tx,n,mx;
struct ZKW{
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
struct segmenttree{int s,a,b;}T[N<<3];
int bs;
ZKW() {bs=1;}
inline void up(int x) {
int tmp=cmin(T[ls(x)].b,T[rs(x)].a);
T[x].s=T[ls(x)].s+T[rs(x)].s+tmp;
T[x].a=T[ls(x)].a+T[rs(x)].a-tmp;
T[x].b=T[ls(x)].b+T[rs(x)].b-tmp;
}
inline void build() {for (;bs<=mx;bs<<=1);}
inline void update(int p,int x,int t) {
p+=bs;
if (t) T[p].b+=x;
else T[p].a+=x;
for (p>>=1;p;p>>=1) up(p);
}
}T;
inline int main() {
//FI=freopen("nanfeng.in","r",stdin);
//FO=freopen("nanfeng.out","w",stdout);
cin >> n;
for (ri i(1);i<=n;p(i)) cin >> bn[i],mx=cmax(mx,bn[i]);
for (ri i(1);i<=n;p(i)) cin >> an[i],mx=cmax(mx,an[i]),p(tmx[an[i]]);
tx=mx;
T.build();
for (ri i(1);i<=n;p(i)) T.update(bn[i],1,1),T.update(an[i],1,0);
int ans=T.T[1].s;
for (ri i(1);i<=n;p(i)) {
T.update(bn[i],-1,1);
ri l=bn[i]+1,r,res(-1);
while(!tmx[tx]) --tx;
r=tx;
while(l<=r) {
int mid(l+r>>1);
T.update(mid,-1,0);
if (T.T[1].s==ans-1) l=mid+1,res=mid;
else r=mid-1;
T.update(mid,1,0);
}
if (res!=-1) --ans,--tmx[res],printf("%d ",res),T.update(res,-1,0);
else {
l=1,r=bn[i],res;
while(l<=r) {
int mid(l+r>>1);
T.update(mid,-1,0);
if (T.T[1].s==ans) l=mid+1,res=mid;
else r=mid-1;
T.update(mid,1,0);
}
T.update(res,-1,0);
printf("%d ",res);
--tmx[res];
}
}
puts("");
return 0;
}
}
int main() {return nanfeng::main();}